Question

Cho a,b,c>0 thỏa mãn \(a^2+b^2+c^2=1\)

Tìm GTNN của A=\(\dfrac{a}{b^2+c^2}+\dfrac{b}{c^2+a^2}+\dfrac{c}{a^2+b^2}\)

(Sử dụng Cauchy)

Answer 1

Áp dụng bất đẳng thức AM-GM:

\(\dfrac{a}{b^2+c^2}+\left(b^2+c^2\right)\ge2\sqrt{a}\)

\(\dfrac{b}{c^2+a^2}+\left(c^2+a^2\right)\ge2\sqrt{b}\)

\(\dfrac{c}{a^2+b^2}+\left(a^2+b^2\right)\ge2\sqrt{c}\)

Cộng theo vế:

\(A+2\ge2\left(\sqrt{a}+\sqrt{b}+\sqrt{c}\right)\)

Mặt khác: \(\left(\sqrt{a}+\sqrt{b}+\sqrt{c}\right)^2\ge3\left(a+b+c\right)\)

\(\left(3a+3b+3c\right)^2\ge27\left(a^2+b^2+c^2\right)=27\)

\(\Rightarrow3\left(a+b+c\right)\ge\sqrt{27}\Leftrightarrow\sqrt{a}+\sqrt{b}+\sqrt{c}\ge\sqrt[4]{27}\)

\(A\ge\sqrt[4]{27}-2\)