Question

Cho a,b,c\(\ge0\),thoa man a+b+c=4.Chung minh rang:

 \(\sqrt{a+b}+\sqrt{b+c}+\sqrt{c+a}\ge4\)

Answer 1

ta có:\(a,b,c\ge0;a+b+c=4\)

\(\Rightarrow a+b\le4\)\(mà\)\(a,b\ge0\)\(\Rightarrow0\le a+b\le4\left(1\right)\)

\(\Rightarrow\sqrt{a+b}\le2\)

\(\Rightarrow2-\sqrt{a+b}\ge0\)\(\left(2\right)\)

Từ (1) và(2)\(\Rightarrow\sqrt{a+b}\left(2-\sqrt{a+b}\right)\ge0\)

\(\Rightarrow2\sqrt{a+b}\ge a+b\)

CMTT:\(2\sqrt{b+c}\ge b+c;2\sqrt{c+a}\ge c+a\)

\(\Rightarrow2\left(\sqrt{a+b}+\sqrt{b+c}+\sqrt{c+a}\right)\ge2\left(a+b+c\right)\)

Mà a+b+c=4\(\Rightarrow\sqrt{a+b}+\sqrt{b+c}+\sqrt{c+a}\ge4\)

Dấu "="xảy ra khi \(\left(a;b;c\right)=\left(4;0;0\right);\left(0;4;0\right);\left(0;0;4\right)\)