Question

Cho các số thực dương a, b, c thỏa mãn a + b + c = 3. Chứng minh rằng:\(\sqrt{a^2+3b^2}+\sqrt{b^2+3c^2}+\sqrt{c^2+3a^2}\ge6\)

Answer 1

Sửa đề \(\sqrt{a^2+bc}+\sqrt{b^2+ca}+\sqrt{c^2+ab}\le6\)

\(\sqrt{a^2+3b}=\sqrt{a^2+\left(a+b+c\right)b}=\sqrt{a^2+ab+b^2+bc}\\ =\sqrt{\left(a+b\right)\left(a+c\right)}\le\dfrac{a+b+a+c}{2}=\dfrac{2a+b+c}{2}\)

Cmtt \(\Leftrightarrow\left\{{}\begin{matrix}\sqrt{b^2+3c}\le\dfrac{a+2b+c}{2}\\\sqrt{c^2+3a}\le\dfrac{a+b+2c}{2}\end{matrix}\right.\)

Cộng VTV:

\(\Leftrightarrow VT\le\dfrac{2a+b+c+a+2b+c+a+b+2c}{2}\\ \Leftrightarrow VT\le\dfrac{4\left(a+b+c\right)}{2}=2\left(a+b+c\right)=6\)

Dấu \("="\Leftrightarrow a=b=c=1\)