Ta có \(\sqrt{a+b}+\sqrt{a-b}\le2\sqrt{a}\)
\(\Leftrightarrow\left(\sqrt{a+b}+\sqrt{a-b}\right)^2\le\left(2\sqrt{a}\right)^2\)\(\Leftrightarrow a+b+a-b+2.\sqrt{\left(a+b\right)\left(a-b\right)}\le4a\)
\(\Leftrightarrow2a+2\sqrt{\left(a+b\right)\left(a-b\right)}\le4a\)
\(\Leftrightarrow-2a+2.\sqrt{\left(a+b\right)\left(a-b\right)}\le0\)\(\Leftrightarrow-\left(2a-2.\sqrt{\left(a+b\right).\left(a-b\right)}\right)\le0\)
\(\Leftrightarrow a+b+a-b-2.\sqrt{\left(a+b\right)\left(a-b\right)}\ge0\)
\(\Leftrightarrow\left(\sqrt{a+b}-\sqrt{a-b}\right)^2\ge0\)( luôn đúng nên suy ra điều phải chứng minh )