Question

Cho 2 số thực ko amm a, b thỏa mãn \(\sqrt{a}+\sqrt{b}=1\)CMR:   \(\text{ ab(a+b)}^2\le\frac{1}{64}\)

Answer 1

\(\sqrt{a}+\sqrt{b}=1\Rightarrow\left(\sqrt{a}+\sqrt{b}\right)^2=1\)

\(\Rightarrow a+b+2\sqrt{ab}=1\)

\(\Rightarrow1-2\sqrt{ab}=a+b\)

Ta có

\(\left(4\sqrt{ab}+1\right)^2\ge0\)

\(\Rightarrow16ab-8\sqrt{ab}+1\ge0\)

\(\Rightarrow8\sqrt{ab}\left(1+2\sqrt{ab}\right)\le1\)

\(\Rightarrow8\sqrt{ab}\left(a+b\right)\le1\)

\(\Rightarrow64ab\left(a+b\right)\le1\)

\(\Rightarrow ab\left(a+b\right)\le\frac{1}{64}\)

(đpcm)