Question

Cho \(a,b>0\)thỏa mãn \(a+b\le1\). Tìm min của \(P=a^2+b^2+\frac{1}{a^2}+\frac{1}{b^2}\)

Answer 1

Áp dụng BĐT AM-GM ta có:

\(ab\le\frac{\left(a+b\right)^2}{4}\le\frac{1}{4}\)

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

\(=a^2+\frac{1}{16a^2}+b^2+\frac{1}{16b^2}+15\left(\frac{1}{16a^2}+\frac{1}{16b^2}\right)\)

\(\ge2\sqrt{a^2\cdot\frac{1}{16a^2}}+2\sqrt{b^2\cdot\frac{1}{16b^2}}+15\cdot2\sqrt{\frac{1}{16a^2}\cdot\frac{1}{16b^2}}\)

\(=\frac{1}{2}+\frac{1}{2}+15\cdot2\cdot\frac{1}{16ab}\)\(\ge1+15\cdot2\cdot\frac{1}{16\cdot\frac{1}{4}}=\frac{17}{2}\)

Xảy ra khi \(a=b=\frac{1}{2}\)