moi nguoi giup em voi a

Question

Cho

$a;b;c>0,$ thỏa mãn:

$9((\frac{1}{a^{2}} + \frac{1}{b^{2}} + \frac{1}{c^{2}}) = 3(\frac{1}{ab} + \frac{1}{bc} + \frac{1}{ac}) + 2014 .$
Tìm GTLN của:
P=

$\frac{1}{\sqrt{5a^{2} + 2ab + 2b^{2}}}$ +

$\frac{1}{\sqrt{5b^{2} + 2bc + 2c^{2}}}$ +

$\frac{1}{\sqrt{5c^{2} + 2ac + 2a^{2}}}$

score 2 · Answer 1

Đặt

$x=\frac{1}{a},y=\frac{1}{b},z=\frac{1}{c}$ .Từ giả thiết có:

$9(x^2+y^2+z^2)=3(xy+yz+xz)+2014\leq 3(x^2+y^2+z^2)+2014\Rightarrow \frac{1007}{3}\geq x^2+y^2+z^2$

Do vậy

$(x+y+z)^2=(x^2+y^2+z^2)+2(xy+yz+xz)\leq 3(x^2+y^2+z^2)\leq 1007$

$5a^2+2ab+2b^2=4a^2+2ab+b^2+(a^2+b^2)\geq 4a^2+4ab+b^2=(2a+b)^2$

$\Rightarrow \frac{1}{\sqrt{5a^2+2ab+2b^2}}\leq \frac{1}{2a+b}\leq \frac{1}{9}(\frac{2}{a}+\frac{1}{b})=\frac{2x+y}{9}$

Tương tự ta có:

$VT\leq \frac{x+y+z}{3}\leq \frac{1}{3}.\sqrt{1007}$

Dấu = xảy ra khi

$x=y=z$

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