Question

cho a>0, b>0 thoả mãn \(\dfrac{1}{a}+\dfrac{1}{b}=2\)

tìm GTLN \(Q=\dfrac{1}{a^4+b^2+2ab^2}+\dfrac{1}{b^4+a^2+2a^2b}\)

Answer 1

Lời giải:

Áp dụng BĐT Cô-si cho các số dương:

\(a^4+b^2\geq 2\sqrt{a^4b^2}=2a^2b\)

\(\Rightarrow a^4+b^2+2ab^2\geq 2a^2b+2ab^2=2ab(a+b)\)

\(\Rightarrow \frac{1}{a^4+b^2+2ab^2}\leq \frac{1}{2ab(a+b)}\)

Tương tự: \(\frac{1}{b^4+a^2+2a^2b}\leq \frac{1}{2ab(a+b)}\)

Do đó: \(Q\leq \frac{1}{2ab(a+b)}+\frac{1}{2ab(a+b)}=\frac{1}{ab(a+b)}\)

Từ đk đầu tiên \(\frac{1}{a}+\frac{1}{b}=2\Leftrightarrow \frac{a+b}{ab}=2\Rightarrow a+b=2ab\)

\(\Rightarrow Q\leq \frac{1}{2a^2b^2}\)

Theo BĐT Cô-si: \(2=\frac{1}{a}+\frac{1}{b}\geq 2\sqrt{\frac{1}{ab}}\Rightarrow ab\geq 1\)

\(\Rightarrow Q\leq \frac{1}{2(ab)^2}\leq \frac{1}{2.1^2}=\frac{1}{2}\)

Vậy \(Q_{\max}=\frac{1}{2}\Leftrightarrow a=b=1\)