Question

a,b,c>0 va a+b+c=1 Chung minh \(\sqrt{a+1}+\sqrt{b+1}+\sqrt{c+1}< 3,5\)

Answer 1

<=> √12.25

<=>a+1+b+1+c+1 12.25

<=>4<12.25(dpcm)

hay

Answer 2

Áp dụng BĐT Bunyakovsky, ta có:

\(\left(a+1+b+1+c+1\right)3\ge\left(\sqrt{a+1}+\sqrt{b+1}+\sqrt{c+1}\right)^2\)

\(\Rightarrow\sqrt{a+1}+\sqrt{b+1}+\sqrt{c+1}\le\sqrt{12}< 3,5\)