Sửa đề: Cho ba số thực a,b,c dương

Áp dụng BĐT Cauchy Schwarz, ta được:

\(VT=\left(a+b+c\right)\left(\frac{9}{bc}+\frac{25}{c+a}+\frac{64}{a+b}\right)-98\ge\left(a+b+c\right)\left(\frac{256}{2\left(a+b+c\right)}\right)-98=30\)

\(\Leftrightarrow VT\ge30\)

Dấu '=' xảy ra khi \(\frac{8}{a+b}=\frac{5}{c+a}=\frac{3}{b+c}\)

\(\Leftrightarrow\frac{8}{a+b}=\frac{8}{a+b+2c}\)

hay c=0(vô lý)

=> Dấu bằng không xảy ra

=>ĐPCM

Đặt \(\hept{\begin{cases}b+c=x>0\\c+a=y>0\\a+b=z>0\end{cases}}\Rightarrow\hept{\begin{cases}a=\frac{y+z-x}{2}\\b=\frac{z+x-y}{2}\\x=\frac{x+y-z}{2}\end{cases}}\)

Bất đẳng thức cần chứng minh tương đương:

\(\frac{9\left(y+z-x\right)}{2x}+\frac{25\left(z+x-y\right)}{2y}+\frac{64\left(x+y-z\right)}{2z}>30\)

Ta có: \(VP=\frac{9y}{2x}+\frac{9z}{2x}-\frac{9}{2}+\frac{25z}{2y}+\frac{25x}{2y}-\frac{9}{2}+\frac{32x}{z}+\frac{32y}{z}-32\)

\(=\left(\frac{9y}{2x}+\frac{25x}{2y}\right)+\left(\frac{9z}{2x}+\frac{32x}{z}\right)+\left(\frac{25z}{2y}+\frac{32y}{z}\right)-41\)

\(\ge2\cdot\frac{15}{2}+2\cdot12+2\cdot20-41=38>30\)

\(\Rightarrow\frac{9a}{b+c}+\frac{25b}{c+a}+\frac{64c}{a+b}>30\)

Đặt \(\left(b+c,c+a,a+b\right)\rightarrow\left(x,y,z\right)\)thì \(x,y,z>0\)và \(a=\frac{y+z-x}{2};b=\frac{z+x-y}{2};c=\frac{x+y-z}{2}\)

Bất đẳng thức cần chứng minh trở thành: \(\frac{y+z-x}{2x}+\frac{25\left(z+x-y\right)}{2y}+\frac{4\left(x+y-z\right)}{2z}>2\)

Xét \(VT=\left(\frac{y}{2x}+\frac{z}{2x}-\frac{1}{2}\right)+\left(\frac{25z}{2y}+\frac{25x}{2y}-\frac{25}{2}\right)+\left(\frac{2x}{z}+\frac{2y}{z}-2\right)\)\(=\left(\frac{y}{2x}+\frac{25x}{2y}\right)+\left(\frac{25z}{2y}+\frac{2y}{z}\right)+\left(\frac{z}{2x}+\frac{2x}{z}\right)-15\)\(\ge2\sqrt{\frac{y}{2x}.\frac{25x}{2y}}+2\sqrt{\frac{25z}{2y}.\frac{2y}{z}}+2\sqrt{\frac{z}{2x}.\frac{2x}{z}}-15=2\)(BĐT Cauchy)

Đẳng thức xảy ra khi \(10x=2y=5z\)hay \(10\left(b+c\right)=2\left(c+a\right)=5\left(a+b\right)\)\(\Rightarrow\hept{\begin{cases}10b+8c=2a\\5b+10c=5a\end{cases}}\Leftrightarrow\hept{\begin{cases}2a=10b+8c\\2a=2b+4c\end{cases}}\Leftrightarrow8b+4c=0\)(Vô lí vì 8b + 4c > 0 với mọi b,c dương)

Vậy dấu bằng không xảy ra

em chao chi a

sao lại chào chị

\(VT=\frac{\left(5a+c\right)^2}{\left(b+c\right)\left(5a+c\right)}+\frac{\left(6b\right)^2}{6b\left(a+c\right)}+\frac{\left(5c+a\right)^2}{\left(a+b\right)\left(5c+a\right)}\)

\(VT\ge\frac{\left(5a+c+6b+5c+a\right)^2}{5ab+5ac+bc+c^2+6ab+6bc+5ac+5bc+a^2+ab}\)

\(VT\ge\frac{36\left(a+b+c\right)^2}{a^2+c^2+12ab+12bc+10ac}\ge\frac{36\left(a+b+c\right)^2}{a^2+c^2+a^2+b^2+b^2+c^2+10ab+10bc+10ac}\)

\(VT\ge\frac{36\left(a+b+c\right)^2}{2\left(a+b+c\right)^2+6\left(ab+bc+ca\right)}\ge\frac{36\left(a+b+c\right)^2}{2\left(a+b+c\right)^2+2\left(a+b+c\right)^2}=9\)

Dấu "=" xảy ra khi \(a=b=c\)