Áp dụng bất đẳng thức cô-si, ta có:
\(\frac{a^2}{b+c}+\frac{b+c}{4}\ge2.\sqrt{\frac{a^2}{b+c}.\frac{b+c}{4}}\)
=>\(\frac{a^2}{b+c}+\frac{b+c}{4}\ge2.\sqrt{\frac{a^2}{4}}\)
=>\(\frac{a^2}{b+c}+\frac{b+c}{4}\ge2.\frac{a}{2}\)
=>\(\frac{a^2}{b+c}+\frac{b+c}{4}\ge a\)
Tương tự, ta có:
\(\frac{b^2}{c+a}+\frac{c+a}{4}\ge b\)
\(\frac{c^2}{a+b}+\frac{a+b}{4}\ge c\)
=>\(\frac{a^2}{b+c}+\frac{b+c}{4}+\frac{b^2}{c+a}+\frac{c+a}{4}+\frac{c^2}{a+b}+\frac{a+b}{4}\ge a+b+c\)
=>\(\frac{a^2}{b+c}+\frac{b^2}{c+a}+\frac{c^2}{a+b}+\left(\frac{b+c}{4}+\frac{c+a}{4}+\frac{a+b}{4}\right)\ge a+b+c\)
=>\(\frac{a^2}{b+c}+\frac{b^2}{c+a}+\frac{c^2}{a+b}+\frac{b+c+c+a+a+b}{4}\ge a+b+c\)
=>\(\frac{a^2}{b+c}+\frac{b^2}{c+a}+\frac{c^2}{a+b}+\frac{a+b+c}{2}\ge\frac{2.\left(a+b+c\right)}{2}\)
=>\(\frac{a^2}{b+c}+\frac{b^2}{c+a}+\frac{c^2}{a+b}\ge\frac{2.\left(a+b+c\right)}{2}-\frac{a+b+c}{2}\)
=>\(\frac{a^2}{b+c}+\frac{b^2}{c+a}+\frac{c^2}{a+b}\ge\frac{a+b+c}{2}\)
Dấu "=" xảy ra khi: \(\frac{a^2}{b+c}=\frac{b+c}{4}=>4.a^2=\left(b+c\right)^2=>2a=b+c=>3a=a+b+c\)
\(\frac{b^2}{c+a}=\frac{c+a}{4}=>4.b^2=\left(c+a\right)^2=>2b=c+a=>3b=a+b+c\)
\(\frac{c^2}{a+b}=\frac{a+b}{4}=>4.c^2=\left(a+b\right)^2=>2c=a+b=>3c=a+b+c\)
=>3a=3b=3c=a+b+c
=>a=b=c
=>ĐPCM
Áp dụng BĐT schwarz:
\(\frac{a^2}{b+c}+\frac{b^2}{c+a}+\frac{c^2}{a+b}\ge\frac{\left(a+b+c\right)^2}{b+c+c+a+a+b}=\frac{\left(a+b+c\right)^2}{2\left(a+b+c\right)}=\frac{a+b+c}{2}\)
Dấu "=" xảy ra khi \(a=b=c\)
