cho a,b,c,d dương và a+b+c+d=1

N

noinhobinhyen

Ta có

$(\sqrt{4a+1}+\sqrt{4b+1}+\sqrt{4c+1}+\sqrt{4d+1})^2 \leq 4(4a+4b+4c+4d+4)=32$

$\Rightarrow \sqrt{4a+1}+\sqrt{4b+1}+\sqrt{4c+1}+\sqrt{4d+1} \leq 4\sqrt{2}$
 
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