Question

Cho a, b > 0. CM: \(\frac{1}{a^2}+\frac{1}{b^2}+\frac{4}{a^2+b^2}\ge\frac{32\left(a^2+b^2\right)}{\left(a+b\right)^4}\)

Answer 1

Lời giải:

Áp dụng BĐT Cauchy:

\(\frac{1}{a^2}+\frac{1}{b^2}+\frac{4}{a^2+b^2}=\frac{a^2+b^2}{a^2b^2}+\frac{4}{a^2+b^2}\geq 2\sqrt{\frac{a^2+b^2}{a^2b^2}.\frac{4}{a^2+b^2}}=\frac{4}{ab}=\frac{32(a^2+b^2)}{8ab(a^2+b^2)}(1)\)

Tiếp tục áp dụng BĐT Cauchy ngược dấu:

\(8ab(a^2+b^2)=4.(2ab).(a^2+b^2)\leq (2ab+a^2+b^2)^2=(a+b)^4(2)\)

Từ \((1);(2)\Rightarrow \frac{1}{a^2}+\frac{1}{b^2}+\frac{4}{a^2+b^2}\geq \frac{32(a^2+b^2)}{8ab(a^2+b^2)}\geq \frac{32(a^2+b^2)}{(a+b)^4}\) (đpcm)

Dấu "=" xảy ra khi $a=b$