Question

Với a; b không âm . chứng minh : \(\frac{\left(a+b\right)^2}{2}+\frac{a+b}{4}\ge a\sqrt{b}+b\sqrt{a}\)

 

Answer 1

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

Áp dụng BĐT cô si 

=> \(\frac{a+b}{2}\ge\sqrt{ab}\)

=> \(\frac{\left(a+b\right)^2}{2}+\frac{a+b}{4}\ge\sqrt{ab}\left(a+b+\frac{1}{2}\right)\) (1)

CM  \(\sqrt{ab}\left(a+b+\frac{1}{2}\right)\ge\) \(a\sqrt{b}+b\sqrt{a}\)

XH : \(\sqrt{ab}\left(a+b+\frac{1}{2}\right)-a\sqrt{b}-b\sqrt{a}\)

= \(\sqrt{ab}\left(a+b+\frac{1}{2}-\sqrt{a}-\sqrt{b}\right)=\sqrt{ab}\left(a-\sqrt{a}+\frac{1}{4}+b-\sqrt{b}+\frac{1}{4}\right)\)

= \(\sqrt{ab}\left[\left(\sqrt{a}-\frac{1}{2}\right)^2+\left(\sqrt{b}-\frac{1}{2}\right)^2\right]\ge0\) Với mọi a ; b > 0 

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