Cho `x,y,z>0` thoã mãn `1/x+1/y+1/z=3`. Tìm GTLN của biểu thức: `P=1/\sqrt{5x^2+2xy+2y^2}+1/\sqrt{5y^2+2yz+2z^2}+1/\sqrt{5z^2+2zx+2x^2}`

Đáp án:

Giải thích các bước giải:

 Ta có:

$5x^2+2xy+2y^2=(4x^2+4xy+y^2)+(x^2-2xy+y^2)=(2x+y)^2+(x-y)^2 \geq (2x+y)^2$

$⇒\dfrac{1}{\sqrt{5x^2+2xy+2y^2}} \leq \dfrac{1}{\sqrt{(2x+y)^2}}=\dfrac{1}{2x+y}$

Mà $\dfrac{1}{2x+y}=\dfrac{1}{x+x+y} \leq \dfrac{1}{9}\left( \dfrac{1}{x}+\dfrac{1}{x}+\dfrac{1}{y}\right)=\dfrac{1}{9}\left( \dfrac{2}{x}+\dfrac{1}{y}\right)$

Do đó:

$\dfrac{1}{\sqrt{5x^2+2xy+2y^2}} \leq \dfrac{1}{9}\left( \dfrac{2}{x}+\dfrac{1}{y}\right)$

Hoàn toàn tương tự:

$\dfrac{1}{\sqrt{5y^2+2yz+2z^2}} \leq \dfrac{1}{9}\left( \dfrac{2}{y}+\dfrac{1}{z}\right)$

$\dfrac{1}{\sqrt{5z^2+2xz+2z^2}} \leq \dfrac{1}{9}\left( \dfrac{2}{z}+\dfrac{1}{x}\right)$

Cộng vế với vế:

$P \leq  \dfrac{1}{9}\left( \dfrac{2}{y}+\dfrac{1}{z}\right)+\dfrac{1}{9}\left( \dfrac{2}{x}+\dfrac{1}{y}\right)+\dfrac{1}{9}\left( \dfrac{2}{z}+\dfrac{1}{x}\right)$

$⇔P \leq \dfrac{1}{9}\left( \dfrac{3}{x}+\dfrac{3}{y}+\dfrac{3}{z}\right)=1$

$P_{max}=1$ khi $x=y=z=1$