\(\sum\)\(\frac{a}{1+a^2}\)\(\le\)\(\sum\)\(\frac{a}{2a}=\frac{3}{2}\)
Dấu "=" xảy ra \(\Leftrightarrow\)\(a=b=c=1\)
\(VT=\frac{a^2}{ab+ca}+\frac{b^2}{bc+ab}+\frac{c^2}{ca+bc}\ge\frac{\left(a+b+c\right)^2}{2\left(ab+bc+ca\right)}\ge\frac{\left(a+b+c\right)^2}{\frac{2}{3}\left(a+b+c\right)^2}=\frac{3}{2}\)
Dấu "=" xảy ra \(\Leftrightarrow\)\(a=b=c\)
sao olm ko hiện \(\sum\) ra nhỉ ? thoi mk ghi lại v
\(\frac{a}{1+a^2}\le\frac{a}{2a}=\frac{1}{2}\)
tương tự 2 cái kia cộng lại t có bđt cần cm
\(\frac{a}{b+c}+\frac{b}{a+c}+\)\(\frac{c}{a+b}\ge\frac{3}{2}\)
Đặt b + c = x
a + c = y
a + b = z
Có: x + y - z = b + c + a + c - a - b = 2c
\(\frac{x+y-z}{2}=c\)
Tương tự: \(\frac{x+z-y}{2}=b\)
\(\frac{z+y-x}{2}=a\)
Khi đó: = \(\frac{z+y-x}{2x}+\frac{x+z-y}{2y}\)\(+\frac{x+y-z}{2z}\)
= \(\frac{z+y}{2x}-\frac{x}{2x}\)\(+\frac{x+z}{2y}-\frac{y}{2y}+\)\(\frac{x+y}{2z}-\frac{z}{2z}\)
= \(\frac{z+y}{2x}-\frac{1}{2}+\frac{x+z}{2y}-\frac{1}{2}\)\(+\frac{x+y}{2z}-\frac{1}{2}\)
= \(\frac{z+y}{2x}+\frac{x+z}{2y}+\frac{x+y}{2z}\)\(-\frac{3}{2}\)
= \(\frac{1}{2}.\left(\frac{z+y}{x}+\frac{x+z}{y}+\frac{x+y}{z}\right)\)\(-\frac{3}{2}\)
= \(\frac{1}{2}.\)\(\left(\frac{z}{x}+\frac{y}{x}+\frac{x}{y}+\frac{z}{y}+\frac{x}{z}+\frac{y}{z}\right)\)\(-\frac{3}{2}\)
Ta có : \(\frac{z}{x}+\frac{x}{z}\ge2\)
\(\frac{y}{x}+\frac{x}{y}\ge2\)
\(\frac{y}{z}+\frac{z}{y}\ge2\)
\(\Rightarrow\)\(\frac{a}{b+c}+\frac{b}{a+c}+\frac{c}{a+b}\ge\)\(\frac{1}{2}.6-\frac{3}{2}\)
\(\frac{a}{b+c}+\frac{b}{a+c}+\frac{c}{a+b}\ge\frac{3}{2}\) ( đpcm )
