\(\frac{\left(2+6a+3b+6\sqrt{2bc}\right)\left(\sqrt{2b^2+2\left(a+c\right)^2}+3\right)}{2a+b+2\sqrt{2bc}}\ge16\)

Ap dung bdt amgm va bdt bunhiacpoxki taok:

\(VT=\frac{\left(2+6a+3b+6\sqrt{2bc}\right)\left(\sqrt{2b^2+2\left(a+c\right)^2}+3\right)}{2a+b+2\sqrt{2bc}}\)

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

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

\(=\left(a+b+c+3\right)\left(\frac{1}{a+b+c}+3\right)\)

\(\ge\left(1+3\right)^2=16=VP\)

dau = khi a+b+c=1;b=a+c;2c=b;a>0;b>0;c>0 tu giai tiep ra

dùng bất đẳng thức cô-si và bất đẳng thức bu-nhi-a-cop-xki

VT=\(\frac{2}{2a+b+2\sqrt{2bc}}+3\)\(\ge\frac{2}{2a+2b+2c}+3\)=\(\frac{1}{a+b+c}+\frac{9}{3}\)

Áp dụng bđt cauchy-schwarzta được:

VT\(\ge\frac{\left(1+3\right)^2}{a+b+c+3}=\frac{6}{a+b+c+3}\)

ta cần chứng minh: a+b+c\(\le\sqrt{2b^2+2\left(a+c\right)^2}\)

Thật vậy ta có

\(2b^2+2\left(a+c\right)^2-\left(a+b+c\right)^2=b^2+a^2+c^2+2ac-2bc-2ab=\left(b-a-c\right)^2\ge0\)Suy ra a+b+c\(\le\sqrt{2b^2+2\left(a+c\right)^2}\)

vậy VT\(\ge VP\)

Mẹ mi đứa nào chả nói được☺️😗😗😎😗😐🤔☺️😉😍😗😄😉