\(\dfrac{a^5-a^2}{a^5+b^2+c^2}+\dfrac{b^5-b^2}{b^5+c^2+a^2}+\dfrac{c^5-c^2}{c^5+a^2+b^2}\ge0\)
\(\Leftrightarrow1-\dfrac{a^2+b^2+c^2}{a^5+b^2+c^2}+1-\dfrac{a^2+b^2+c^2}{b^5+c^2+a^2}+1-\dfrac{a^2+b^2+c^2}{c^5+a^2+b^2}\ge0\)
\(\Leftrightarrow\dfrac{1}{a^5+b^2+c^2}+\dfrac{1}{b^5+c^2+a^2}+\dfrac{1}{c^5+a^2+b^2}\le\dfrac{3}{a^2+b^2+c^2}\)
Áp dụng BĐT Cauchy-Schwarz ta có:
\(\left(a^5+b^2+c^2\right)\left(\dfrac{1}{a}+b^2+c^2\right)\ge\left(a^2+b^2+c^2\right)^2\)
\(\Rightarrow\dfrac{1}{a^5+b^2+c^2}\le\dfrac{\dfrac{1}{a}+b^2+c^2}{\left(a^2+b^2+c^2\right)^2}\). Tương tự cho 2 BĐT còn lại rồi cộng theo vế:
\(VT\le\dfrac{\dfrac{1}{a}+\dfrac{1}{b}+\dfrac{1}{c}+2\left(a^2+b^2+c^2\right)}{\left(a^2+b^2+c^2\right)^2}\)
Cần cm \(\dfrac{1}{a}+\dfrac{1}{b}+\dfrac{1}{c}+2\left(a^2+b^2+c^2\right)\le3\left(a^2+b^2+c^2\right)\)
\(\Leftrightarrow\dfrac{1}{a}+\dfrac{1}{b}+\dfrac{1}{c}\le a^2+b^2+c^2\Leftrightarrow\dfrac{ab+bc+ca}{abc}\le a^2+b^2+c^2\)
\(\Leftrightarrow ab+bc+ca\le a^2+b^2+c^2\) *ĐÚNG*
