Question

[Yim Yim]

\(cho:a;b\ge0.CMR:\)

\(\sqrt{a+b}\le\sqrt{a}+\sqrt{b}\le\sqrt{2\left(a+b\right)}\)

Answer 1

[HeroZombie]

*)\(\sqrt{a}+\sqrt{b}\ge\sqrt{a+b}\)

\(\Leftrightarrow a+b+2\sqrt{ab}\ge a+b\)

\(\Leftrightarrow2\sqrt{ab}\ge0\) (luôn đúng)

*)\(\sqrt{2\left(a+b\right)}\ge\sqrt{a}+\sqrt{b}\)

\(\Leftrightarrow2\left(a+b\right)\ge a+b+2\sqrt{ab}\)

\(\Leftrightarrow a+b\ge2\sqrt{ab}\)\(\Leftrightarrow\left(a+b\right)^2\ge4ab\)

\(\Leftrightarrow a^2-2ab+b^2\ge0\Leftrightarrow\left(a-b\right)^2\ge0\) (luôn đúng, too)

Answer 2

[HeroZombie]

*)\(\sqrt{a}+\sqrt{b}\ge\sqrt{a+b}\)

\(\Leftrightarrow a+b+2\sqrt{ab}\ge a+b\)

\(\Leftrightarrow2\sqrt{ab}\ge0\) (luôn đúng)

*)\(\sqrt{2\left(a+b\right)}\ge\sqrt{a}+\sqrt{b}\)

\(\Leftrightarrow2\left(a+b\right)\ge a+b+2\sqrt{ab}\)

\(\Leftrightarrow a+b\ge2\sqrt{ab}\)\(\Leftrightarrow\left(a+b\right)^2\ge4ab\)

\(\Leftrightarrow a^2-2ab+b^2\ge0\Leftrightarrow\left(a-b\right)^2\ge0\) (luôn đúng, too)