Theo giả thiết, ta có: \(a^2b^2+b^2c^2+c^2a^2=a^2b^2c^2\Leftrightarrow\frac{1}{a^2}+\frac{1}{b^2}+\frac{1}{c^2}=1\)
Áp dụng BĐT AM - GM cho 5 số, ta được: \(\hept{\begin{cases}a.a.a.b.b\le\frac{a^5+a^5+a^5+b^5+b^5}{5}=\frac{3a^5+2b^5}{5}\\b.b.b.a.a\le\frac{b^5+b^5+b^5+a^5+a^5}{5}=\frac{3b^5+2a^5}{5}\end{cases}}\)
\(\Rightarrow\frac{5\left(a^5+b^5\right)}{5}\ge a^2b^2\left(a+b\right)\)hay \(a^5+b^5\ge a^2b^2\left(a+b\right)\)
\(\Rightarrow\frac{1}{\sqrt{a^5+b^5}}\le\frac{1}{ab\sqrt{a+b}}\)(1) .
Tương tự, ta có: \(\frac{1}{\sqrt{b^5+c^5}}\le\frac{1}{bc\sqrt{b+c}}\)(2); \(\frac{1}{\sqrt{c^5+a^5}}\le\frac{1}{ca\sqrt{c+a}}\)(3)
Cộng theo vế của 3 BĐT (1), (2), (3), ta được: \(VT=\Sigma_{cyc}\frac{1}{\sqrt{a^5+b^5}}\le\Sigma_{cyc}\frac{1}{ab\sqrt{a+b}}\)()
Xét \(\left(\Sigma_{cyc}\frac{1}{ab\sqrt{a+b}}\right)^2\le\left(\frac{1}{a^2}+\frac{1}{b^2}+\frac{1}{c^2}\right)\left(\Sigma_{cyc}\frac{1}{b^2\left(a+b\right)}\right)\)\(=\Sigma_{cyc}\frac{1}{b^2\left(a+b\right)}\Rightarrow\Sigma_{cyc}\frac{1}{ab\sqrt{a+b}}\le\sqrt{\Sigma_{cyc}\frac{1}{b^2\left(a+b\right)}}\)(2)
Từ (1) và (2) suy ra \(\Sigma_{cyc}\frac{1}{\sqrt{a^5+b^5}}\le\sqrt{\Sigma_{cyc}\frac{1}{b^2\left(a+b\right)}}\)(đpcm)
Đẳng thức xảy ra khi \(a=b=c=\sqrt{3}\)
