Question

[Nguyễn Thị Phương Linh]

1 . cho a,b,c >0 thỏa mãn 2ab+6bc+2ac=7abc

tìm GTNN \(M=\frac{4ab}{a+2b}+\frac{9ac}{a+4c}+\frac{4bc}{b+c}\)

2 . Cho x , y , z > 0 và x+y+z=1

Chứng minh rằng : \(\frac{1-x^2}{x+yz}+\frac{1-y^2}{y+xz}+\frac{1-z^2}{z+xy}\ge6\)

Answer 1

[Agatsuma Zenitsu]

Bài 1: Theo đề : \(2ab+6bc+2ac=7abc\) \(;a,b,c>0\)

Chia cả 2 vế cho \(abc>0\Rightarrow\frac{2}{c}+\frac{6}{a}+\frac{2}{b}=7\)

Đặt: \(\hept{\begin{cases}x=\frac{1}{a}\\y=\frac{1}{b}\\z=\frac{1}{c}\end{cases}}\Rightarrow\hept{\begin{cases}x,y,z>0\\2z+6x+2y=7\end{cases}}\)

Khi đó: \(M=\frac{4ab}{a+2b}+\frac{9ac}{a+4c}+\frac{4bc}{b+c}=\frac{4}{2x+y}+\frac{9}{4x+z}+\frac{4}{y+z}\)

\(\Rightarrow M=\frac{4}{2x+y}+2x+y+\frac{9}{4x+z}+4x+z+\frac{4}{y+z}+y+z-\left(2x+y+4x+z+y+z\right)\)

\(=\left(\frac{2}{\sqrt{x+2y}}-\sqrt{x+2y}\right)^2+\left(\frac{3}{\sqrt{4x+z}}-\sqrt{4x+z}\right)^2+\left(\frac{2}{\sqrt{y+z}}-\sqrt{y+z}\right)^2+17\ge17\)

Khi: \(\hept{\begin{cases}x=\frac{1}{2}\\y=z=1\end{cases}}\Rightarrow M=17\)

\(Min_M=17\Leftrightarrow a=2;b=1;c=1\)

Answer 2

[tth_new]

ミ★๖ۣۜBăηɠ ๖ۣۜBăηɠ ★彡 chém bài khó nhất rồi nên em xin mạn phép chém bài dễ ạ.

2/\(VT=\Sigma_{cyc}\frac{\left(x+y+z\right)^2-x^2}{x\left(x+y+z\right)+yz}=\Sigma_{cyc}\frac{\left(y+z\right)\left(2x+y+z\right)}{\left(x+y\right)\left(x+z\right)}\)

\(\ge\Sigma_{cyc}\frac{\left(y+z\right)\left(2x+y+z\right)}{\frac{\left(2x+y+z\right)^2}{4}}=\Sigma_{cyc}\frac{4\left(y+z\right)}{2x+y+z}=\Sigma_{cyc}\frac{2\left(y+z-2x\right)}{2x+y+z}+6\)

\(=\Sigma_{cyc}\left(\frac{2\left(x+y+z\right)\left(y+z-2x\right)}{2x+y+z}-\frac{3}{2}\left(y+z-2x\right)\right)+6\)

\(=\Sigma_{cyc}\frac{\left(y+z-2x\right)^2}{2\left(2x+y+z\right)}+6\ge6\)