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sửa đổi
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BĐT!!!
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C2) ta có $3(x^{2}+y^{2}+z^{2}\geq (x+y+z)^{2} \Rightarrow x+y+z \leq \sqrt{3(x^{2}+y^{2}+z^{2}}$ $\Rightarrow P\leq \frac{xyz(\sqrt{3(x^{2}+y^{2}+z^{2}}+\sqrt{x^{2}+y^{2}+z^{2}})}{(x^{2}+y^{2}+z^{2})(xy+yz+zx)}$ $=\frac{(\sqrt{3}+1)xyz}{(xy+yz+zx)(\sqrt{x^{2}+y^{2}+z^{2}})}$ mà $(xy+yz+zx)(\sqrt{x^{2}+y^{2}+z^{2}} \geq 3\sqrt[3]{x^{2}y^{2}z^{2}}\sqrt{3\sqrt[3]{x^{2}y^{2}z^{2}}}=3\sqrt{3}xyz$ $\Rightarrow P\leq \frac{\sqrt{3}+1}{3\sqrt{3}}=\frac{3+\sqrt{3}}{9}$ dấu '=" $\Leftrightarrow a=b=c$
C2) ta có $3(x^{2}+y^{2}+z^{2})\geq (x+y+z)^{2} \Rightarrow x+y+z \leq \sqrt{3(x^{2}+y^{2}+z^{2})}$ $\Rightarrow P\leq \frac{xyz(\sqrt{3(x^{2}+y^{2}+z^{2}}+\sqrt{x^{2}+y^{2}+z^{2}})}{(x^{2}+y^{2}+z^{2})2(xy+yz+zx)}$ =$\frac{(\sqrt{3}+1)xyz}{2(xy+yz+zx)\sqrt{x^{2}+y^{2}+z^{2}}}$ mà $(xy+yz+zx)\sqrt{x^{2}+y^{2}+z^{2}} \geq 3\sqrt[3]{x^{2}y^{2}z^{2}}\sqrt{3\sqrt[3]{x^{2}y^{2}z^{2}}}=3\sqrt{3}xyz$ $\Rightarrow P\leq \frac{\sqrt{3}+1}{2.3\sqrt{3}}=\frac{3+\sqrt{3}}{18}$ dấu '=" $\Leftrightarrow a=b=c$
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giải đáp
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BĐT!!!
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C2) ta có $3(x^{2}+y^{2}+z^{2})\geq (x+y+z)^{2} \Rightarrow x+y+z \leq \sqrt{3(x^{2}+y^{2}+z^{2})}$ $\Rightarrow P\leq \frac{xyz(\sqrt{3(x^{2}+y^{2}+z^{2}}+\sqrt{x^{2}+y^{2}+z^{2}})}{(x^{2}+y^{2}+z^{2})2(xy+yz+zx)}$ =$\frac{(\sqrt{3}+1)xyz}{2(xy+yz+zx)\sqrt{x^{2}+y^{2}+z^{2}}}$ mà $(xy+yz+zx)\sqrt{x^{2}+y^{2}+z^{2}} \geq 3\sqrt[3]{x^{2}y^{2}z^{2}}\sqrt{3\sqrt[3]{x^{2}y^{2}z^{2}}}=3\sqrt{3}xyz$ $\Rightarrow P\leq \frac{\sqrt{3}+1}{2.3\sqrt{3}}=\frac{3+\sqrt{3}}{18}$ dấu '=" $\Leftrightarrow a=b=c$ |
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sửa đổi
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........................BĐT............................
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C2)bđt $\Leftrightarrow \frac{\sqrt[3]{\frac{a}{b}}}+\sqrt[3]{\frac{b}{c}}+\sqrt[3]{\frac{c}{a}}}{\sqrt[3]{3(a+b+c)(\frac{1}{a}+\frac{1}{b}+\frac{1}{c})}} \leq1$ $\Leftrightarrow \sum \frac{\sqrt[3]{\frac{a}{b}}}{\sqrt[3]{3(a+b+c)(\frac{1}{a}+ \frac{1}{b} +\frac{1}{c})}}\leq1$ ta có $\frac{\sqrt[3]{\frac{a}{b}}}{\sqrt[3]{3(a+b+c)(\frac{1}{a}+frac{1}{b}+\frac{1}{c})}} =\sqrt[3]{\frac{1}{3}.\frac{a}{a+b+c}.\frac{\frac{1}{b}}{\frac{1}{a}+\frac{1}{b}+\frac{1}{c})}}$ $\leq \frac{1}{3}(\frac{1}{3}+\frac{a}{a+b+c}+\frac{\frac{1}{b}}{\frac{1}{a}+\frac{1}{b}+\frac{1}{c}})$ TT $\Rightarrow VT\leq \frac{1}{3}(1+1+1)=1$ dấu '=" $\Leftrightarrow a=b=c$
C2)bđt $\Leftrightarrow \frac{\sqrt[3]{\frac{a}{b}}+\sqrt[3]{\frac{b}{c}}+\sqrt[3]{\frac{c}{a}}}{\sqrt[3]{3(a+b+c)(\frac{1}{a}+\frac{1}{b}+\frac{1}{c})}} \leq1$ $\Leftrightarrow \sum \frac{\sqrt[3]{\frac{a}{b}}}{\sqrt[3]{3(a+b+c)(\frac{1}{a}+ \frac{1}{b} +\frac{1}{c})}}\leq1$ ta có $\frac{\sqrt[3]{\frac{a}{b}}}{\sqrt[3]{3(a+b+c)(\frac{1}{a}+\frac{1}{b}+\frac{1}{c})}} =\sqrt[3]{\frac{1}{3}.\frac{a}{a+b+c}.\frac{\frac{1}{b}}{\frac{1}{a}+\frac{1}{b}+\frac{1}{c})}}$ $\leq \frac{1}{3}(\frac{1}{3}+\frac{a}{a+b+c}+\frac{\frac{1}{b}}{\frac{1}{a}+\frac{1}{b}+\frac{1}{c}})$ TT $\Rightarrow VT\leq \frac{1}{3}(1+1+1)=1$ dấu '=" $\Leftrightarrow a=b=c$
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sửa đổi
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........................BĐT............................
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C2)bđt $\Leftrightarrow \frac{\sqrt[3]{\frac{a}{b}+\sqrt[3]{\frac{b}{c}}+\sqrt[3]{\frac{c}{a}}}}{\sqrt[3]{3(a+b+c)(\frac{1}{a}+\frac{1}{b}+\frac{1}{c})}} \leq1$ $\Leftrightarrow \sum \frac{\sqrt[3]{\frac{a}{b}}}{\sqrt[3]{3(a+b+c)(\frac{1}{a}+\frac{1}{b}+\frac{1}{c})}}\leq1$ ta có $\frac{\sqrt[3]{\frac{a}{b}}}{\sqrt[3]{3(a+b+c)(\frac{1}{a}+frac{1}{b}+\frac{1}{c})}} =\sqrt[3]{\frac{1}{3}.\frac{a}{a+b+c}.\frac{\frac{1}{b}}{\frac{1}{a}+\frac{1}{b}+\frac{1}{c})}}$ $\leq \frac{1}{3}(\frac{1}{3}+\frac{a}{a+b+c}+\frac{\frac{1}{b}}{\frac{1}{a}+\frac{1}{b}+\frac{1}{c}})$ TT $\Rightarrow VT\leq \frac{1}{3}(1+1+1)=1$ dấu '=" $\Leftrightarrow a=b=c$
C2)bđt $\Leftrightarrow \frac{\sqrt[3]{\frac{a}{b}}}+\sqrt[3]{\frac{b}{c}}+\sqrt[3]{\frac{c}{a}}}{\sqrt[3]{3(a+b+c)(\frac{1}{a}+\frac{1}{b}+\frac{1}{c})}} \leq1$ $\Leftrightarrow \sum \frac{\sqrt[3]{\frac{a}{b}}}{\sqrt[3]{3(a+b+c)(\frac{1}{a}+ \frac{1}{b} +\frac{1}{c})}}\leq1$ ta có $\frac{\sqrt[3]{\frac{a}{b}}}{\sqrt[3]{3(a+b+c)(\frac{1}{a}+frac{1}{b}+\frac{1}{c})}} =\sqrt[3]{\frac{1}{3}.\frac{a}{a+b+c}.\frac{\frac{1}{b}}{\frac{1}{a}+\frac{1}{b}+\frac{1}{c})}}$ $\leq \frac{1}{3}(\frac{1}{3}+\frac{a}{a+b+c}+\frac{\frac{1}{b}}{\frac{1}{a}+\frac{1}{b}+\frac{1}{c}})$ TT $\Rightarrow VT\leq \frac{1}{3}(1+1+1)=1$ dấu '=" $\Leftrightarrow a=b=c$
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giải đáp
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........................BĐT............................
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C2) bđt $\Leftrightarrow \frac{\sqrt[3]{\frac{a}{b}}+\sqrt[3]{\frac{b}{c}}+\sqrt[3]{\frac{c}{a}}}{\sqrt[3]{3(a+b+c)(\frac{1}{a}+\frac{1}{b}+\frac{1}{c})}} \leq1$ $\Leftrightarrow \sum \frac{\sqrt[3]{\frac{a}{b}}}{\sqrt[3]{3(a+b+c)(\frac{1}{a}+ \frac{1}{b} +\frac{1}{c})}}\leq1$ ta có $\frac{\sqrt[3]{\frac{a}{b}}}{\sqrt[3]{3(a+b+c)(\frac{1}{a}+\frac{1}{b}+\frac{1}{c})}} =\sqrt[3]{\frac{1}{3}.\frac{a}{a+b+c}.\frac{\frac{1}{b}}{\frac{1}{a}+\frac{1}{b}+\frac{1}{c})}}$ $\leq \frac{1}{3}(\frac{1}{3}+\frac{a}{a+b+c}+\frac{\frac{1}{b}}{\frac{1}{a}+\frac{1}{b}+\frac{1}{c}})$ TT $\Rightarrow VT\leq \frac{1}{3}(1+1+1)=1$ dấu '=" $\Leftrightarrow a=b=c$ |
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được thưởng
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Đăng nhập hàng ngày 12/06/2016
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được thưởng
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Đăng nhập hàng ngày 11/06/2016
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giải đáp
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help với
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10. Q≤12a2b+2ab2+12ab2+2a2b=1ab(a+b) 1a+1b=2⇒a+b=2ab ⇒Q≤12a2b2 (a+b)2≥4ab⇔(a+b)2≥2(a+b)⇔a+b≥2 ⇔a+bab≥2ab⇔2≥2ab ⇒ab≥1 ⇒Q≤12 dấu "=" ⇔a=b=1 |
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bình luận
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Chán ngán -_-
t cũng k nhớ rõ tại cứ vào tìm lung tung thấy là đọc thui
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bình luận
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Chán ngán -_-
thôi nha bà. tui dọc 1 lần oi siêu j có bà ý!!!
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sửa đổi
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Chán ngán -_-
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ta có $x\sqrt{8y-5}\leq \sqrt{3} x\sqrt{\frac{8y-5}{3}} \leq \frac{\sqrt{3}}{2}x(\frac{8y-5}{3}+1)=\frac{\sqrt{3}}{3}x(4y-1)$ $\Rightarrow VT(1) \leq \frac{\sqrt{3}}{3}(8xy-x-y) \leq \frac{\sqrt{3}}{3}(2(x+y)^{2}-x-y)$ VP $\geq \sqrt[4]{24(\frac{1}{2}((x+y)^{2}+4))} =\sqrt[4]{12(x+y)^{2}+96}$ ta cần cm $\sqrt[4]{12(x+y)^{2}+96}\geq \frac{\sqrt{3}}{3} (2(x+y)^{2}-x-y)$ đặt $t=x+y$ $\Leftrightarrow \sqrt[4]{12t^{2}+96}\geq \frac{\sqrt{3}}{3}(2t^{2}-1)$ nâng lũy thừa bậc 4 $\Leftrightarrow (t-2) ( t^{7}+\frac{3t}{2}^{5}+\frac{5t}{2}^{4}+\frac{8t}{16}^{3}+\frac{81t}{8}^{2}+\frac{27t}{2}+2)\leq 0$ $\Leftrightarrow t\leq 2$ ta cần tìm đk này từ pt (2) vs t-x=y (2) $\Leftrightarrow 11x^{2}-6x(t-x)+3(t-x)=12x-4(t-x)$ $\Leftrightarrow 20x^{2}-(16+12t)x+3t^{2}+4=0$ $\Delta' \geq 0 \Leftrightarrow t\in \left[ \frac{-4}{3}{;}2 \right]$ $\Rightarrow x=y=1$
ta có $x\sqrt{8y-5}\leq \sqrt{3} x\sqrt{\frac{8y-5}{3}} \leq \frac{\sqrt{3}}{2}x(\frac{8y-5}{3}+1)=\frac{\sqrt{3}}{3}x(4y-1)$ $\Rightarrow VT(1) \leq \frac{\sqrt{3}}{3}(8xy-x-y) \leq \frac{\sqrt{3}}{3}(2(x+y)^{2}-x-y)$ VP $\geq \sqrt[4]{24(\frac{1}{2}((x+y)^{2}+4))} =\sqrt[4]{12(x+y)^{2}+96}$ ta cần cm $\sqrt[4]{12(x+y)^{2}+96}\geq \frac{\sqrt{3}}{3} (2(x+y)^{2}-x-y)$ đặt $t=x+y$ $\Leftrightarrow \sqrt[4]{12t^{2}+96}\geq \frac{\sqrt{3}}{3}(2t^{2}-1)$ nâng lũy thừa bậc 4 $\Leftrightarrow (t-2) ( t^{7}+\frac{3t}{2}^{5}+\frac{5t}{2}^{4}+\frac{8t}{16}^{3}+\frac{81t}{8}^{2}+\frac{27t}{2}+2)\leq 0$ $\Leftrightarrow t\leq 2$ ta cần tìm đk này từ pt (2) vs t-x=y (2) $\Leftrightarrow 11x^{2}-6x(t-x)+3(t-x)^{2}=12x-4(t-x)$ $\Leftrightarrow 20x^{2}-(16+12t)x+3t^{2}+4t=0$ $\Delta' \geq 0 \Leftrightarrow t\in \left[ \frac{-4}{3}{;}2 \right]$ $\Rightarrow x=y=1$
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sửa đổi
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Chán ngán -_-
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ta có $x\sqrt{8y-5}\leq \sqrt{3} x\sqrt{\frac{8y-5}{3}} \leq \frac{\sqrt{3}}{2}x(\frac{8y-5}{3}+1)=\frac{\sqrt{3}}{3}x(4y-1)$ $\Rightarrow VT(1) \leq \frac{\sqrt{3}}{3}(8xy-x-y) \leq \frac{\sqrt{3}}{3}(2(x+y)^{2}-x-y)$ VP $\geq \sqrt[4]{24(\frac{1}{2}((x+y)^{2}+4))} =\sqrt[4]{12(x+y)^{2}+96}$ ta cần cm $\sqrt[4]{12(x+y)^{2}+96}\geq \frac{\sqrt{3}}{3} (2(x+y)^{2}-x-y)$ đặt $t=x+y$ $\Leftrightarrow \sqrt[4]{12t^{2}+96}\geq \frac{\sqrt{3}}{3}(2t^{2}-1)$ nâng lũy thừa bậc 4 $\Leftrightarrow (t-2) ( t^{7}+\frac{3t}{2}^{5}+\frac{5t}{2}^{4}+\frac{8t}{16}^{3}+\frac{81t}{8}^{2}+\frac{27t}{2}+2)\leq 0$ $\Leftrightarrow t\leq 2$ ta cần tìm đk này từ pt (2) vs t-x=y (2) $\Leftrightarrow 11x^{2}-6x(t-x)+3(t-x)=12x-4(t-x)$ $\Leftrightarrow 20x^{2}-(16+12y)x+3t^{2}+4=0$ $\Delta' \geq 0 \Leftrightarrow t\in \left[ \frac{-4}{3}{;}2 \right]$ $\Rightarrow x=y=1$
ta có $x\sqrt{8y-5}\leq \sqrt{3} x\sqrt{\frac{8y-5}{3}} \leq \frac{\sqrt{3}}{2}x(\frac{8y-5}{3}+1)=\frac{\sqrt{3}}{3}x(4y-1)$ $\Rightarrow VT(1) \leq \frac{\sqrt{3}}{3}(8xy-x-y) \leq \frac{\sqrt{3}}{3}(2(x+y)^{2}-x-y)$ VP $\geq \sqrt[4]{24(\frac{1}{2}((x+y)^{2}+4))} =\sqrt[4]{12(x+y)^{2}+96}$ ta cần cm $\sqrt[4]{12(x+y)^{2}+96}\geq \frac{\sqrt{3}}{3} (2(x+y)^{2}-x-y)$ đặt $t=x+y$ $\Leftrightarrow \sqrt[4]{12t^{2}+96}\geq \frac{\sqrt{3}}{3}(2t^{2}-1)$ nâng lũy thừa bậc 4 $\Leftrightarrow (t-2) ( t^{7}+\frac{3t}{2}^{5}+\frac{5t}{2}^{4}+\frac{8t}{16}^{3}+\frac{81t}{8}^{2}+\frac{27t}{2}+2)\leq 0$ $\Leftrightarrow t\leq 2$ ta cần tìm đk này từ pt (2) vs t-x=y (2) $\Leftrightarrow 11x^{2}-6x(t-x)+3(t-x)=12x-4(t-x)$ $\Leftrightarrow 20x^{2}-(16+12t)x+3t^{2}+4=0$ $\Delta' \geq 0 \Leftrightarrow t\in \left[ \frac{-4}{3}{;}2 \right]$ $\Rightarrow x=y=1$
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