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sửa đổi
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làm phép tính rõ ràng cho cái
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làm phép tính rõ ràng cho cái Tinh tổng $1+2+3...+n$
làm phép tính rõ ràng cho cái Ch o N$\g eq$1, n$\in$NCMR: $1+2+3 +...+n$ $=$$\frac{n(n+1)}{2}$
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sửa đổi
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giai phuong trinh day
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giai phuong trinh day Giải phương trình :a,$\frac{1}{a+b-x}$ = $\frac{1}{a}$$+$$\frac{1}{b}$$+$$\frac{1}{x}$(x là ẩn số)b,$\frac{(b-c)(1+a^2)}{x+a^2}$ + $\frac{(c-a)(1+b^2)}{x+b^2}$+$\frac{(a-b)(1+c^2)}{x+c^2}$$($a,b,c là hằng số đôi một khác nhau)
giai phuong trinh day Giải phương trình :a,$\frac{1}{a+b-x}$ = $\frac{1}{a}$$+$$\frac{1}{b}$$+$$\frac{1}{x}$(x là ẩn số)b,$\frac{(b-c)(1+a^2)}{x+a^2}$ + $\frac{(c-a)(1+b^2)}{x+b^2}$+$\frac{(a-b)(1+c^2)}{x+c^2}$$ =$$0$$($a,b,c là hằng số đôi một khác nhau)
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sửa đổi
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giup em voi
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giup em voi Cho ba số thực không âm a,b,c thỏa mãn $a b$$+$$b c$$+$$c a$$=$$1$. Chứng minh rằng $:$1$\leq$$\frac{a}{1+bc}$$+$$\frac{b}{1+ca}$$+$$\frac{c}{1+ab}$$\leq $ $\frac{3\sqrt{3}}{4}$
giup em voi Cho ba số thực không âm a,b,c thỏa mãn $a ^2$$+$$b ^2$$+$$c ^2$$=$$1$. Chứng minh rằng $:$1$\leq$$\frac{a}{1+bc}$$+$$\frac{b}{1+ca}$$+$$\frac{c}{1+ab}$$\leq $ $\frac{3\sqrt{3}}{4}$
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sửa đổi
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giup em voi
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giup em voi Cho ba số thực không âm a,b,c thỏa mãn $a ^2$$+$$b ^2$$+$$c ^2$$=$$1$. Chứng minh rằng $:$1$\leq$$\frac{a}{1+bc}$$+$$\frac{b}{1+ca}$$+$$\frac{c}{1+ab}$$\leq $ $\frac{3\sqrt{3}}{4}$
giup em voi Cho ba số thực không âm a,b,c thỏa mãn $a b$$+$$b c$$+$$c a$$=$$1$. Chứng minh rằng $:$1$\leq$$\frac{a}{1+bc}$$+$$\frac{b}{1+ca}$$+$$\frac{c}{1+ab}$$\leq $ $\frac{3\sqrt{3}}{4}$
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sửa đổi
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giúp em mấy bài này với,
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Ta có : $\frac{a}{b+2c+3d}$ = $\frac{a^2}{ab+2ac+3ad}$Tương tự ..........cộng lại $\frac{a}{b+2c+3d}$ + $\frac{b}{c+2d+3a}$ +$\frac{c}{d+2a+3b}$ + $\frac{d}{a+2b+3c}$= $\frac{a^2}{ab+2ac+3ad}$ + $\frac{b^2}{bc+2db+3ab}$ + .............>= $\frac{(a+b+c+d)^2}{4(ab+ac+ad+bc+bd+cd}$Ta có $\frac{3}{2}$. (a+b+c+d)^2 - 4(ab+bc+ca+ad+bd+cd) = $\frac{3}{2}$.$(a^2+b^2+c^2+d^2+2ab+2bc+2cd+2da+2ac+2bd)$ - $4(ab+bc+ca+ad+bd+cd)$=$\frac{3(a^2+b^2+c^2+d^2)-2(ab+bc+ca+ad+bd+cd)}{2}$=$\frac{(a-b)^2+(b-c)^2+(c-a)^2+(a-d)^2+(b-d)^2}{2}$ $\geq $0 $\Rightarrow$ $4(ab+bc+ca+ad+bd+cd)$ $\leq$ $\frac{3}{2}$ .$(a+b+c+d)^2$$\Rightarrow$ dpcm
Ta có : $\frac{a}{b+2c+3d}$ = $\frac{a^2}{ab+2ac+3ad}$Tương tự ..........cộng lại $\frac{a}{b+2c+3d}$ + $\frac{b}{c+2d+3a}$ +$\frac{c}{d+2a+3b}$ + $\frac{d}{a+2b+3c}$= $\frac{a^2}{ab+2ac+3ad}$ + $\frac{b^2}{bc+2db+3ab}$ + .............>= $\frac{(a+b+c+d)^2}{4(ab+ac+ad+bc+bd+cd}$Ta có $\frac{3}{2}$. (a+b+c+d)^2 - 4(ab+bc+ca+ad+bd+cd) = $\frac{3}{2}$.$(a^2+b^2+c^2+d^2+2ab+2bc+2cd+2da+2ac+2bd)$ - $4(ab+bc+ca+ad+bd+cd)$=$\frac{3(a^2+b^2+c^2+d^2)-2(ab+bc+ca+ad+bd+cd)}{2}$=$\frac{(a-b)^2+(b-c)^2+(c-a)^2+(a-d)^2+(b-d)^2}{2}$ $\geq $0 $\Rightarrow$ 4$($ab$+$bc$+ca+ad+bd+cd)$ $\leq$ $\frac{3}{2}$ .$(a+b+c+d)^2$$\Rightarrow$ dpcm
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sửa đổi
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help me
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Ta có : $\frac{a^2}{a+2b^3}$ = a - $\frac{2ab^3}{a+2b^3}$ $\geq$ a - $\frac{2ab^3}{3\sqrt[3]{ab^6}}$ = a - $\frac{2b\sqrt[3]{a^2} }{3}$Tương tự : $\frac{b^2}{b+2c^3}$ $\geq$ b - $\frac{2c\sqrt[3]{b^2}}{3}$ ; $\frac{c^2}{c+2a^3}$ $\geq$ c - $\frac{2a\sqrt[3]{c^2}}{3}$$\Rightarrow$ $\frac{a^2}{a+2b^3}$ + $\frac{b^2}{b+2c^3}$ + $\frac{c^2}{c+2a^3}$ $\geq $ a+b+c - $\frac{2}{3}$.$( b\sqrt[3]{a^2}$ +$c\sqrt[3]{b^2}$ + $a\sqrt[3]{b^3}$ )Ta cần chứng minh : a+b+c - $\frac{2}{3}$ . $( b\sqrt[3]{a^2}$ + $c\sqrt[3]{b^2}$ + $a\sqrt[3]{b^3}$ ) $\geq $1 $\Leftrightarrow $3 - $\frac{2}{3}$ . $( b\sqrt[3]{a^2}$ + $c\sqrt[3]{b^2}$ + $a\sqrt[3]{b^3}$ ) $\geq$ 1 $\Leftrightarrow$ $\frac{2}{3}$ . $( b\sqrt[3]{a^2}$ + $c\sqrt[3]{b^2}$ + $a\sqrt[3]{b^3}$ ) $\leq$ 3Ta có $b\sqrt[3]{a^2}$ $\leq$ $\frac{b.(2a+1)}{3}$Tương tự rồi cộng lại $\Rightarrow$ dpcm
Ta có : $\frac{a^2}{a+2b^3}$ = a - $\frac{2ab^3}{a+2b^3}$ $\geq$ a - $\frac{2ab^3}{3\sqrt[3]{ab^6}}$ = a - $\frac{2b\sqrt[3]{a^2} }{3}$Tương tự : $\frac{b^2}{b+2c^3}$ $\geq$ b - $\frac{2c\sqrt[3]{b^2}}{3}$ ; $\frac{c^2}{c+2a^3}$ $\geq$ c - $\frac{2a\sqrt[3]{c^2}}{3}$$\Rightarrow$ $\frac{a^2}{a+2b^3}$ + $\frac{b^2}{b+2c^3}$ + $\frac{c^2}{c+2a^3}$ $\geq $ a+b+c - $\frac{2}{3}$.$( b\sqrt[3]{a^2}$ +$c\sqrt[3]{b^2}$ + $a\sqrt[3]{b^3}$ )Ta cần chứng minh : a+b+c - $\frac{2}{3}$ . $( b\sqrt[3]{a^2}$ + $c\sqrt[3]{b^2}$ + $a\sqrt[3]{b^3}$ ) $\geq $1 $\Leftrightarrow $3 - $\frac{2}{3}$ . $( b\sqrt[3]{a^2}$ + $c\sqrt[3]{b^2}$ + $a\sqrt[3]{b^3}$ ) $\geq$ 1 $\Leftrightarrow$ $( b\sqrt[3]{a^2}$ + $c\sqrt[3]{b^2}$ + $a\sqrt[3]{b^3}$ ) $\leq$ 3Ta có $b\sqrt[3]{a^2}$ $\leq$ $\frac{b.(2a+1)}{3}$Tương tự rồi cộng lại $\Rightarrow$ dpcm
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sửa đổi
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giúp em mấy bài này với,
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Ta có : $\frac{a}{b+2c+3d}$ = $\frac{a^2}{ab+2ac+3ad}$Tương tự ..........cộng lại $\frac{a}{b+2c+3d}$ + $\frac{b}{c+2d+3a}$ +$\frac{c}{d+2a+3b}$ + $\frac{d}{a+2b+3c}$= $\frac{a^2}{ab+2ac+3ad}$ + $\frac{b^2}{bc+2db+3ab}$ + .............>= $\frac{(a+b+c+d)^2}{4(ab+ac+ad+bc+bd+cd}$Ta có $\frac{3}{2}$. (a+b+c+d)^2 - 4(ab+bc+ca+ad+bd+cd) = $\frac{3}{2}$.(a^2+b^2+c^2+d^2+2ab+2bc+2cd+2da+2ac+2bd)-4(ab+bc+ca+ad+bd+cd)=$\frac{3(a^2+b^2+c^2+d^2)-2(ab+bc+ca+ad+bd+cd)}{2}$=$\frac{(a-b)^2+(b-c)^2+(c-a)^2+(a-d)^2+(b-d)^2}{2}$ $\geq $0 $\Rightarrow$ 4(ab+bc+ca+ad+bd+cd) $\leq$ $\frac{3}{2}$ .(a+b+c+d)^2$\Rightarrow$ dpcm
Ta có : $\frac{a}{b+2c+3d}$ = $\frac{a^2}{ab+2ac+3ad}$Tương tự ..........cộng lại $\frac{a}{b+2c+3d}$ + $\frac{b}{c+2d+3a}$ +$\frac{c}{d+2a+3b}$ + $\frac{d}{a+2b+3c}$= $\frac{a^2}{ab+2ac+3ad}$ + $\frac{b^2}{bc+2db+3ab}$ + .............>= $\frac{(a+b+c+d)^2}{4(ab+ac+ad+bc+bd+cd}$Ta có $\frac{3}{2}$. (a+b+c+d)^2 - 4(ab+bc+ca+ad+bd+cd) = $\frac{3}{2}$.$(a^2+b^2+c^2+d^2+2ab+2bc+2cd+2da+2ac+2bd)$ - $4(ab+bc+ca+ad+bd+cd)$=$\frac{3(a^2+b^2+c^2+d^2)-2(ab+bc+ca+ad+bd+cd)}{2}$=$\frac{(a-b)^2+(b-c)^2+(c-a)^2+(a-d)^2+(b-d)^2}{2}$ $\geq $0 $\Rightarrow$ $4(ab+bc+ca+ad+bd+cd)$ $\leq$ $\frac{3}{2}$ .$(a+b+c+d)^2$$\Rightarrow$ dpcm
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sửa đổi
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giúp em mấy bài này với,
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Ta có : $\frac{a}{b+2c+3d}$ = $\frac{a^2}{ab+2ac+3ad}$Tương tự ..........cộng lại $\frac{a}{b+2c+3d}$ + $\frac{b}{c+2d+3a}$ +$\frac{c}{d+2a+3b}$ + $\frac{d}{a+2b+3c}$= $\frac{a^2}{ab+2ac+3ad}$ + $\frac{b^2}{bc+2db+3ab}$ + .............>= $\frac{(a+b+c+d)^2}{4(ab+ac+ad+bc+bd+cd}$Ta có $\frac{3}{2}$. (a+b+c+d)^2 - 4(ab+bc+ca+ad+bd+cd) = $\frac{3}{2}$.(a^2+b^2+c^2+d^2+2ab+2bc+2cd+2da+2ac+2bd)-4(ab+bc+ca+ad+bd+cd)=$\frac{3(a^2+b^2+c^2+d^2)-2(ab+bc+ca+ad+bd+cd)}{2}$=$\frac{(a-b)^2+(b-c)^2+(c-a)^2+(a-d)^2+(b-d)^2}{2}$ $\geq $0 $\Rightarrow$ 4(ab+bc+ca+ad+bd+cd) $\leq$ $\frac{3}{2}$ .(a+b+c+d)^2$\Rightarrow$ dpcm
Ta có : $\frac{a}{b+2c+3d}$ = $\frac{a^2}{ab+2ac+3ad}$Tương tự ..........cộng lại $\frac{a}{b+2c+3d}$ + $\frac{b}{c+2d+3a}$ +$\frac{c}{d+2a+3b}$ + $\frac{d}{a+2b+3c}$= $\frac{a^2}{ab+2ac+3ad}$ + $\frac{b^2}{bc+2db+3ab}$ + .............>= $\frac{(a+b+c+d)^2}{4(ab+ac+ad+bc+bd+cd}$Ta có $\frac{3}{2}$. (a+b+c+d)^2 - 4(ab+bc+ca+ad+bd+cd) = $\frac{3}{2}$.(a^2+b^2+c^2+d^2+2ab+2bc+2cd+2da+2ac+2bd)-4(ab+bc+ca+ad+bd+cd)=$\frac{3(a^2+b^2+c^2+d^2)-2(ab+bc+ca+ad+bd+cd)}{2}$=$\frac{(a-b)^2+(b-c)^2+(c-a)^2+(a-d)^2+(b-d)^2}{2}$ $\geq $0 $\Rightarrow$ 4(ab+bc+ca+ad+bd+cd) $\leq$ $\frac{3}{2}$ .(a+b+c+d)^2$\Rightarrow$ dpcm
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sửa đổi
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giúp em mấy bài này với,
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Ta có : $\frac{a}{b+2c+3d}$ = $\frac{a^2}{ab+2ac+3ad}$Tương tự ..........cộng lại $\frac{a}{b+2c+3d}$ + $\frac{b}{c+2d+3a}$ + $\frac{c}d+2a+3b}$ + $\frac{d}{a+2b+3c}$ = $\frac{a^2}{ab+2ac+3ad}$ + $\frac{b^2}{bc+2db+3ab}$ + .............>= $\frac{(a+b+c+d)^2}{4(ab+ac+ad+bc+bd+cd}$Ta có $\frac{3}{2}$. (a+b+c+d)^2 - 4(ab+bc+ca+ad+bd+cd) = $\frac{3}{2}$.(a^2+b^2+c^2+d^2+2ab+2bc+2cd+2da+2ac+2bd)-4(ab+bc+ca+ad+bd+cd)=$\frac{3(a^2+b^2+c^2+d^2)-2(ab+bc+ca+ad+bd+cd)}{2}$=$\frac{(a-b)^2+(b-c)^2+(c-a)^2+(a-d)^2+(b-d)^2}{2}$ $\geq $0 $\Rightarrow$ 4(ab+bc+ca+ad+bd+cd) $\leq $ $\frac{3}{2}$ .(a+b+c+d)^2$\Rightarrow$ dpcm
Ta có : $\frac{a}{b+2c+3d}$ = $\frac{a^2}{ab+2ac+3ad}$Tương tự ..........cộng lại $\frac{a}{b+2c+3d}$ + $\frac{b}{c+2d+3a}$ +$\frac{c}{d+2a+3b}$ + $\frac{d}{a+2b+3c}$= $\frac{a^2}{ab+2ac+3ad}$ + $\frac{b^2}{bc+2db+3ab}$ + .............>= $\frac{(a+b+c+d)^2}{4(ab+ac+ad+bc+bd+cd}$Ta có $\frac{3}{2}$. (a+b+c+d)^2 - 4(ab+bc+ca+ad+bd+cd) = $\frac{3}{2}$.(a^2+b^2+c^2+d^2+2ab+2bc+2cd+2da+2ac+2bd)-4(ab+bc+ca+ad+bd+cd)=$\frac{3(a^2+b^2+c^2+d^2)-2(ab+bc+ca+ad+bd+cd)}{2}$=$\frac{(a-b)^2+(b-c)^2+(c-a)^2+(a-d)^2+(b-d)^2}{2}$ $\geq $0 $\Rightarrow$ 4(ab+bc+ca+ad+bd+cd) $\leq$ $\frac{3}{2}$ .(a+b+c+d)^2$\Rightarrow$ dpcm
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sửa đổi
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giúp em mấy bài này với,
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Ta có : $\frac{a}{b+2c+3d}$ = $\frac{a^2}{ab+2ac+3ad}$Tương tự ..........cộng lại $\frac{a}{b+2c+3d}$ + $\frac{b}{c+2d+3a}$ + $\frac{c}{d+2a+3b}$ + $\frac{d}{a+2b+3c}$ = $\frac{a^2}{ab+2ac+3ad}$ + $\frac{b^2}{bc+2db+3ab}$ + .............>= $\frac{(a+b+c+d)^2}{4(ab+ac+ad+bc+bd+cd}$Ta có $\frac{3}{2}$. (a+b+c+d)^2 - 4(ab+bc+ca+ad+bd+cd) = $\frac{3}{2}$.(a^2+b^2+c^2+d^2+2ab+2bc+2cd+2da+2ac+2bd)-4(ab+bc+ca+ad+bd+cd)=$\frac{3(a^2+b^2+c^2+d^2)-2(ab+bc+ca+ad+bd+cd)}{2}$=$\frac{(a-b)^2+(b-c)^2+(c-a)^2+(a-d)^2+(b-d)^2}{2}$ $\geq $0 $\Rightarrow$ 4(ab+bc+ca+ad+bd+cd) $\leq $ $\frac{3}{2}$ .(a+b+c+d)^2$\Rightarrow$ dpcm
Ta có : $\frac{a}{b+2c+3d}$ = $\frac{a^2}{ab+2ac+3ad}$Tương tự ..........cộng lại $\frac{a}{b+2c+3d}$ + $\frac{b}{c+2d+3a}$ + $\frac{c}d+2a+3b}$ + $\frac{d}{a+2b+3c}$ = $\frac{a^2}{ab+2ac+3ad}$ + $\frac{b^2}{bc+2db+3ab}$ + .............>= $\frac{(a+b+c+d)^2}{4(ab+ac+ad+bc+bd+cd}$Ta có $\frac{3}{2}$. (a+b+c+d)^2 - 4(ab+bc+ca+ad+bd+cd) = $\frac{3}{2}$.(a^2+b^2+c^2+d^2+2ab+2bc+2cd+2da+2ac+2bd)-4(ab+bc+ca+ad+bd+cd)=$\frac{3(a^2+b^2+c^2+d^2)-2(ab+bc+ca+ad+bd+cd)}{2}$=$\frac{(a-b)^2+(b-c)^2+(c-a)^2+(a-d)^2+(b-d)^2}{2}$ $\geq $0 $\Rightarrow$ 4(ab+bc+ca+ad+bd+cd) $\leq $ $\frac{3}{2}$ .(a+b+c+d)^2$\Rightarrow$ dpcm
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sửa đổi
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giúp em mấy bài này với,
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Ta có : \frac{a}{b+2c+3d} = \frac{a^2}{ab+2ac+3ad}Tương tự ..........cộng lại \frac{a}{b+2c+3d} + \frac{b}{c+2d+3a} + \frac{c}{d+2a+3b} + \frac{d}{a+2b+3c} = \frac{a^2}{ab+2ac+3ad} + \frac{b^2}{bc+2db+3ab} + .............>= \frac{(a+b+c+d)^2}{4(ab+ac+ad+bc+bd+cd}Ta có \frac{3}{2}. (a+b+c+d)^2 - 4(ab+bc+ca+ad+bd+cd) = \frac{3}{2}.(a^2+b^2+c^2+d^2+2ab+2bc+2cd+2da+2ac+2bd)-4(ab+bc+ca+ad+bd+cd)=\frac{3(a^2+b^2+c^2+d^2)-2(ab+bc+ca+ad+bd+cd)}{2}=\frac{(a-b)^2+(b-c)^2+(c-a)^2+(a-d)^2+(b-d)^2}{2} >= 0 \Rightarrow 4(ab+bc+ca+ad+bd+cd) <= \frac{3}{2} .(a+b+c+d)^2\Rightarrow dpcm
Ta có : $\frac{a}{b+2c+3d}$ = $\frac{a^2}{ab+2ac+3ad}$Tương tự ..........cộng lại $\frac{a}{b+2c+3d}$ + $\frac{b}{c+2d+3a}$ + $\frac{c}{d+2a+3b}$ + $\frac{d}{a+2b+3c}$ = $\frac{a^2}{ab+2ac+3ad}$ + $\frac{b^2}{bc+2db+3ab}$ + .............>= $\frac{(a+b+c+d)^2}{4(ab+ac+ad+bc+bd+cd}$Ta có $\frac{3}{2}$. (a+b+c+d)^2 - 4(ab+bc+ca+ad+bd+cd) = $\frac{3}{2}$.(a^2+b^2+c^2+d^2+2ab+2bc+2cd+2da+2ac+2bd)-4(ab+bc+ca+ad+bd+cd)=$\frac{3(a^2+b^2+c^2+d^2)-2(ab+bc+ca+ad+bd+cd)}{2}$=$\frac{(a-b)^2+(b-c)^2+(c-a)^2+(a-d)^2+(b-d)^2}{2}$ $\geq $0 $\Rightarrow$ 4(ab+bc+ca+ad+bd+cd) $\leq $ $\frac{3}{2}$ .(a+b+c+d)^2$\Rightarrow$ dpcm
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sửa đổi
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help me
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Ta có : $\frac{a^2}{a+2b^3}$ = a - $\frac{2ab^3}{a+2b^3}$ $\geq$ a - $\frac{2ab^3}{3\sqrt[3]{ab^6}}$ = a - $\frac{2b\sqrt[3]{a^2} }{3}$Tương tự : $\frac{b^2}{b+2c^3}$ $\geq$ b - $\frac{2c\sqrt[3]{b^2}}{3}$ ; $\frac{c^2}{c+2a^3}$ $\geq$ c - $\frac{2a\sqrt[3]{c^2}}{3}$$\Rightarrow$ $\frac{a^2}{a+2b^3}$ + $\frac{b^2}{b+2c^3}$ + $\frac{c^2}{c+2a^3}$ $\geq $ a+b+c - $\frac{2}{3}$.$( b\sqrt[3]{a^2}$ +$c\sqrt[3]{b^2}$ + $a\sqrt[3]{b^3}$ )Ta cần chứng minh : a+b+c - $\frac{2}{3}$.( $b\sqrt[3]{a^2}$ +$c\sqrt[3]{b^2}$ + $a\sqrt[3]{b^3}$ ) $\geq$ 1 \Leftrightarrow 3 - $\frac{2}{3}$.( $b\sqrt[3]{a^2}$ +$c\sqrt[3]{b^2}$ + $a\sqrt[3]{b^3}$ ) \geq 1 $$\Leftrightarrow$ $\frac{2}{3}$ .$( b\sqrt[3]{a^2}$ + $c\sqrt[3]{b^2}$ +$ a\sqrt[3]{b^3}$ $\leq$ 3Ta có $b\sqrt[3]{a^2}$ $\leq$ $\frac{b.(2a+1)}{3}$Tương tự rồi cộng lại $\Rightarrow$ dpcm
Ta có : $\frac{a^2}{a+2b^3}$ = a - $\frac{2ab^3}{a+2b^3}$ $\geq$ a - $\frac{2ab^3}{3\sqrt[3]{ab^6}}$ = a - $\frac{2b\sqrt[3]{a^2} }{3}$Tương tự : $\frac{b^2}{b+2c^3}$ $\geq$ b - $\frac{2c\sqrt[3]{b^2}}{3}$ ; $\frac{c^2}{c+2a^3}$ $\geq$ c - $\frac{2a\sqrt[3]{c^2}}{3}$$\Rightarrow$ $\frac{a^2}{a+2b^3}$ + $\frac{b^2}{b+2c^3}$ + $\frac{c^2}{c+2a^3}$ $\geq $ a+b+c - $\frac{2}{3}$.$( b\sqrt[3]{a^2}$ +$c\sqrt[3]{b^2}$ + $a\sqrt[3]{b^3}$ )Ta cần chứng minh : a+b+c - $\frac{2}{3}$ . $( b\sqrt[3]{a^2}$ + $c\sqrt[3]{b^2}$ + $a\sqrt[3]{b^3}$ ) $\geq $1 $\Leftrightarrow $3 - $\frac{2}{3}$ . $( b\sqrt[3]{a^2}$ + $c\sqrt[3]{b^2}$ + $a\sqrt[3]{b^3}$ ) $\geq$ 1 $\Leftrightarrow$ $\frac{2}{3}$ . $( b\sqrt[3]{a^2}$ + $c\sqrt[3]{b^2}$ + $a\sqrt[3]{b^3}$ ) $\leq$ 3Ta có $b\sqrt[3]{a^2}$ $\leq$ $\frac{b.(2a+1)}{3}$Tương tự rồi cộng lại $\Rightarrow$ dpcm
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sửa đổi
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help me
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Ta có : $\frac{a^2}{a+2b^3}$ = a - $\frac{2ab^3}{a+2b^3}$ $\geq$ a - $\frac{2ab^3}{3\sqrt[3]{ab^6}}$ = a - $\frac{2b\sqrt[3]{a^2} }{3}$Tương tự : $\frac{b^2}{b+2c^3}$ $\geq$ b - $\frac{2c\sqrt[3]{b^2}}{3}$ ; $\frac{c^2}{c+2a^3}$ $\geq$ c - $\frac{2a\sqrt[3]{c^2}}{3}$$\Rightarrow$ $\frac{a^2}{a+2b^3}$ + $\frac{b^2}{b+2c^3}$ + $\frac{c^2}{c+2a^3}$ $\geq$ a+b+c - $\frac{2}{3}.( b\sqrt[3]{a^2} +c\sqrt[3]{b^2} + a\sqrt[3]{b^3} )Ta cần chứng minh : a+b+c - \frac{2}{3}.( b\sqrt[3]{a^2} +c\sqrt[3]{b^2} + a\sqrt[3]{b^3} ) \geq 1 \Leftrightarrow 3 - \frac{2}{3}.( b\sqrt[3]{a^2} +c\sqrt[3]{b^2} + a\sqrt[3]{b^3} ) \geq 1 \Leftrightarrow \frac{2}{3}.( b\sqrt[3]{a^2} +c\sqrt[3]{b^2} + a\sqrt[3]{b^3} \leq 3Ta có b\sqrt[3]{a^2} \leq \frac{b.(2a+1)}{3}Tương tự rồi cộng lại \Rightarrow dpcm
Ta có : $\frac{a^2}{a+2b^3}$ = a - $\frac{2ab^3}{a+2b^3}$ $\geq$ a - $\frac{2ab^3}{3\sqrt[3]{ab^6}}$ = a - $\frac{2b\sqrt[3]{a^2} }{3}$Tương tự : $\frac{b^2}{b+2c^3}$ $\geq$ b - $\frac{2c\sqrt[3]{b^2}}{3}$ ; $\frac{c^2}{c+2a^3}$ $\geq$ c - $\frac{2a\sqrt[3]{c^2}}{3}$$\Rightarrow$ $\frac{a^2}{a+2b^3}$ + $\frac{b^2}{b+2c^3}$ + $\frac{c^2}{c+2a^3}$ $\geq $ a+b+c - $\frac{2}{3}$.$( b\sqrt[3]{a^2}$ +$c\sqrt[3]{b^2}$ + $a\sqrt[3]{b^3}$ )Ta cần chứng minh : a+b+c - $\frac{2}{3}$.( $b\sqrt[3]{a^2}$ +$c\sqrt[3]{b^2}$ + $a\sqrt[3]{b^3}$ ) $\geq$ 1 \Leftrightarrow 3 - $\frac{2}{3}$.( $b\sqrt[3]{a^2}$ +$c\sqrt[3]{b^2}$ + $a\sqrt[3]{b^3}$ ) \geq 1 $$\Leftrightarrow$ $\frac{2}{3}$ .$( b\sqrt[3]{a^2}$ + $c\sqrt[3]{b^2}$ +$ a\sqrt[3]{b^3}$ $\leq$ 3Ta có $b\sqrt[3]{a^2}$ $\leq$ $\frac{b.(2a+1)}{3}$Tương tự rồi cộng lại $\Rightarrow$ dpcm
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Ta có : \frac{a^2}{a+2b^3} = a - \frac{2ab^3}{a+2b^3} \geq a - \frac{2ab^3}{3\sqrt[3]{ab^6}} = a - \frac{2b\sqrt[3]{a^2} }{3}Tương tự : \frac{b^2}{b+2c^3} \geq b - \frac{2c\sqrt[3]{b^2}}{3} ; \frac{c^2}{c+2a^3} \geq c - \frac{2a\sqrt[3]{c^2}}{3}\Rightarrow \frac{a^2}{a+2b^3} + \frac{b^2}{b+2c^3} + \frac{c^2}{c+2a^3} \geq a+b+c - \frac{2}{3}.( b\sqrt[3]{a^2} +c\sqrt[3]{b^2} + a\sqrt[3]{b^3} )Ta cần chứng minh : a+b+c - \frac{2}{3}.( b\sqrt[3]{a^2} +c\sqrt[3]{b^2} + a\sqrt[3]{b^3} ) \geq 1 \Leftrightarrow 3 - \frac{2}{3}.( b\sqrt[3]{a^2} +c\sqrt[3]{b^2} + a\sqrt[3]{b^3} ) \geq 1 \Leftrightarrow \frac{2}{3}.( b\sqrt[3]{a^2} +c\sqrt[3]{b^2} + a\sqrt[3]{b^3} \leq 3Ta có b\sqrt[3]{a^2} \leq \frac{b.(2a+1)}{3}Tương tự rồi cộng lại \Rightarrow dpcm
Ta có : $\frac{a^2}{a+2b^3}$ = a - $\frac{2ab^3}{a+2b^3}$ $\geq$ a - $\frac{2ab^3}{3\sqrt[3]{ab^6}}$ = a - $\frac{2b\sqrt[3]{a^2} }{3}$Tương tự : $\frac{b^2}{b+2c^3}$ $\geq$ b - $\frac{2c\sqrt[3]{b^2}}{3}$ ; $\frac{c^2}{c+2a^3}$ $\geq$ c - $\frac{2a\sqrt[3]{c^2}}{3}$$\Rightarrow$ $\frac{a^2}{a+2b^3}$ + $\frac{b^2}{b+2c^3}$ + $\frac{c^2}{c+2a^3}$ $\geq$ a+b+c - $\frac{2}{3}.( b\sqrt[3]{a^2} +c\sqrt[3]{b^2} + a\sqrt[3]{b^3} )Ta cần chứng minh : a+b+c - \frac{2}{3}.( b\sqrt[3]{a^2} +c\sqrt[3]{b^2} + a\sqrt[3]{b^3} ) \geq 1 \Leftrightarrow 3 - \frac{2}{3}.( b\sqrt[3]{a^2} +c\sqrt[3]{b^2} + a\sqrt[3]{b^3} ) \geq 1 \Leftrightarrow \frac{2}{3}.( b\sqrt[3]{a^2} +c\sqrt[3]{b^2} + a\sqrt[3]{b^3} \leq 3Ta có b\sqrt[3]{a^2} \leq \frac{b.(2a+1)}{3}Tương tự rồi cộng lại \Rightarrow dpcm
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help me Cho các số dương $a,b,c$ thỏa mãn điều kiện a$+b+c=3$.Chung minh rang:$a^2/(a+2b^3) + b^2/(b+2c^3) +c^2/(c+2a^3) >=1$ $\frac{a^2}{a+2b}$+$\frac{b^2}{b+2c}$+$\frac{c^2}{c+2a}$ $\geq$ 1
help me Cho các số dương $a,b,c$ thỏa mãn điều kiện a$+b+c=3$.Chung minh rang:$a^2/(a+2b^3) + b^2/(b+2c^3) +c^2/(c+2a^3) >=1$
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