Question

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

Cmr \(\dfrac{1}{a^2+b^2}+\dfrac{1}{b^2+c^2}+\dfrac{1}{c^2+a^2}\le\dfrac{a^3+b^3+c^3}{2abc}+3\)

Answer 1

Lời giải:

Vì $1=a^2+b^2+c^2$ nên:

\(\frac{1}{a^2+b^2}+\frac{1}{b^2+c^2}+\frac{1}{c^2+a^2}=\frac{a^2+b^2+c^2}{a^2+b^2}+\frac{a^2+b^2+c^2}{b^2+c^2}+\frac{a^2+b^2+c^2}{c^2+a^2}\)

\(=3+\frac{a^2}{b^2+c^2}+\frac{b^2}{c^2+a^2}+\frac{c^2}{a^2+b^2}\)

\(\leq 3+\frac{a^2}{2bc}+\frac{b^2}{2ac}+\frac{c^2}{2ab}\) (theo BĐT AM-GM)

\(=3+\frac{a^3+b^3+c^3}{2abc}\) (đpcm)

Dấu "=" xảy ra khi \(a=b=c=\sqrt{\frac{1}{3}}\)