Question

[quang phan duy]

cho a,b,c>0 CMR a^2/a+bc + b^2/b+ac + c^2/c+ab >=a+b+c/4

Answer 1

[Phương Trâm]

Bài này thiếu đề. Đề đúng là phải có \(\frac{1}{a}+\frac{1}{b}+\frac{1}{c}=1\) nữa nha bạn.

\(\frac{a^2}{a+bc}+\frac{b^2}{b+ac}+\frac{c^2}{c+ab}\ge\frac{a+b+c}{4}\)

Ta có: \(\frac{1}{a}+\frac{1}{b}+\frac{1}{c}=1\) \(\Rightarrow ab+bc+ac=abc\)

\(VT=\frac{a^2}{a+bc}+\frac{b^2}{b+ac}+\frac{c^2}{c+ab}\)

\(\Rightarrow VT=\frac{a^2.a}{a\left(a+bc\right)}+\frac{b^2.b}{b\left(b+ac\right)}+\frac{c^2.c}{c\left(c+ab\right)}\)

\(\Leftrightarrow VT=\frac{a^3}{a^2+abc}+\frac{b^3}{b^2+abc}+\frac{c^3}{c^2+abc}\)

\(\Leftrightarrow VT=\frac{a^3}{a^2+ab+bc+ac}+\frac{b^3}{b^2+ab+bc+ac}+\frac{c^3}{c^2+ab+bc+ac}\)

\(\Leftrightarrow VT=\frac{a^3}{a\left(a+b\right)+c\left(a+b\right)}+\frac{b^3}{a\left(b+c\right)+b\left(b+c\right)}+\frac{c^3}{c\left(b+c\right)+a\left(b+c\right)}\)

\(\Leftrightarrow VT=\frac{a^3}{\left(a+c\right)\left(a+b\right)}+\frac{b^3}{\left(b+c\right)\left(a+b\right)}+\frac{c^3}{\left(b+c\right)\left(a+c\right)}\)

Áp dụng BĐT Cauchy ta có:

\(\frac{a^3}{\left(a+b\right)\left(a+c\right)}+\frac{a+b}{8}+\frac{a+c}{8}\ge3\sqrt[3]{\frac{a^3}{64}}=\frac{3a}{4}\)

\(\frac{b^3}{\left(a+b\right)\left(b+c\right)}+\frac{a+b}{8}+\frac{b+c}{8}\ge3\sqrt[3]{\frac{b^3}{64}}=\frac{3b}{4}\)

\(\frac{c^3}{\left(b+c\right)\left(a+c\right)}+\frac{b+c}{8}+\frac{a+c}{8}\ge3\sqrt[3]{\frac{c^3}{64}}=\frac{3c}{4}\)

Ta có:

\(\frac{3a}{4}+\frac{3b}{4}+\frac{3c}{4}+\frac{a+b+c}{2}\ge\frac{3}{4}\left(a+b+c\right)\)

\(\Rightarrow\frac{3a}{4}+\frac{3b}{4}+\frac{3c}{4}\ge\frac{3}{4}\left(a+b+c\right)-\frac{1}{2}\left(a+b+c\right)\)

\(\Rightarrow VT\ge\frac{a+b+c}{4}=VP\)

Dấu \("="\) xảy ra \(\Leftrightarrow a=b=c=3\)

\(\RightarrowĐpcm.\)