ta có$:\sqrt{c(a-c)}+\sqrt{c(b-c)}=\sqrt{c}.\sqrt{a-c}+\sqrt{c}.\sqrt{b-c}(*)$$A/D:BCS ,$ta có$:(*)^2\leq (c+b-c)(a-c+c)=ab\Rightarrow đpcm$
nếu bên ngoài là $c^2$ thì lm như sau:ta có$:\sqrt{c(a-c)}+\sqrt{c(b-c)}=\sqrt{c}.\sqrt{a-c}+\sqrt{c}.\sqrt{b-c}(*)$$A/D:BCS ,$ta có$:(*)^2\leq (c+b-c)(a-c+c)=ab\Rightarrow đpcm$
ta có$:\sqrt{c(a-c)}+\sqrt{c(b-c)}=\sqrt{c}.\sqrt{a-c}+\sqrt{c}.\sqrt{b-c}(*)$$A/D:BCS ,$ta có$:(*)^2\leq (c+b-c)(a-c+c)=ab\Rightarrow đpcm$