## xhejni 2 years ago if a>0 and b>0 prove sqrt of a +sqrt of b >sqrt of a+b

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1. vf321

For $$a,b>0$$, $2\sqrt{ab}>0$Thus: $a+2\sqrt{ab}+b>a+b$Let $$\sqrt{a}$$ and $$v=\sqrt{b}$$. Then,$u^2+2uv+v^2>u^2+v^2$$(u+v)^2>u^2+v^2$$\sqrt{(u+v)^2}=\sqrt{u^2+v^2}$$|u+v|=\sqrt{u^2+v^2}$Since $$u,v>0$$, $$|u+v|=u+v$$. Replace $$u$$ and $$v$$ with $$a$$ and $$b$$.

2. vf321

*appropriate expressions for a and b