Bài toán: Cho $a,b,c>0$ thỏa mãn $15\left( {\dfrac{1}{{{a^2}}} + \dfrac{1}{{{b^2}}} + \dfrac{1}{{{c^2}}}} \right) = 10\left( {\dfrac{1}{{ab}} + \dfrac{1}{{bc}} + \dfrac{1}{{ca}}} \right) + 2011$ Tìm giá trị lớn nhất của
$P=\dfrac{1}{\sqrt{5a^2+2ab+2b^2}}+\dfrac{1}{\sqrt{5b^2+2bc+2c^2}}+\dfrac{1}{\sqrt{5c^2+2ac+2a^2}}$
Lời giải:     Đặt $x=\dfrac{1}{a};y=\dfrac{1}{b};z=\dfrac{1}{c}\\  \rightarrow 15\left(x^{2}+y^{2}+z^{2} \right)=10\left(xy+yz+zx \right)+2011\leq 10\left(x^{2}+y^{2}+z^{2} \right)+2011 \\ \rightarrow x^{2}+y^{2}+z^{2}\leq \dfrac{2011}{5}$.
Mặt khác $\left(x+y+z \right)^{2}\leq 3\left(x^{2}+y^{2}+z^{2} \right)\rightarrow \left(x+y+z \right)^{2}\leq 3.\dfrac{2011}{5} $
Ta có $\dfrac{1}{\sqrt{5a^{2}+2ab+b^{2}}}\leq \dfrac{1}{\sqrt{4a^{2}+2ab+b^{2}+2ab}}=\dfrac{1}{2a+b}\leq \dfrac{1}{9}\left(\dfrac{2}{a}+\dfrac{1}{b} \right)=\dfrac{1}{9}\left(2x+y \right)$
Tương tự ta có $P\leq \dfrac{1}{3}\left(x+y+z \right)\leq \dfrac{1}{3}\sqrt{3.\dfrac{2011}{5}}$