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Áp dụng bđt Svacxơ ta có : VT >= (a+b+c)^2/(2a+2b+2c) = (a+b+c)/2 = VP
=> đpcm
Áp dụng BDT Bunhiacopxki:
\(\left[\left(\sqrt{x+y}\right)^2+\left(\sqrt{y+z}\right)^2+\left(\sqrt{x+z}\right)^2\right]\left[\frac{x^2}{\left(\sqrt{x+y}\right)^2}+\frac{y^2}{\left(\sqrt{y+z}\right)^2}+\frac{z^2}{\left(\sqrt{x+z}\right)^2}\right]\)\(\ge\left(x+y+z\right)^2\)
\(\Leftrightarrow2\left(x+y+z\right)\left(\frac{x^2}{x+y}+\frac{y^2}{y+z}+\frac{z^2}{x+z}\right)\ge\left(x+y+z\right)^2\)
\(\Leftrightarrow\frac{x^2}{x+y}+\frac{y^2}{y+z}+\frac{z^2}{x+z}\ge\frac{x+y+z}{2}\)
\(\sum\dfrac{a}{b^2+bc+c^2}\ge\dfrac{\left(a+b+c\right)^2}{ab^2+abc+ac^2+bc^2+abc+ba^2+ca^2+abc+cb^2}=\dfrac{\left(a+b+c\right)^2}{\left(a+b+c\right)\left(ab+bc+ac\right)}=\dfrac{a+b+c}{ab+bc+ac}\)
Áp dụng BĐT Cauchy Sshwarz, ta có:
\(\frac{a^2}{b+c}+\frac{b^2}{c+a}+\frac{c^2}{a+b}\ge\frac{\left(a+b+c\right)^2}{2\left(a+b+c\right)}=\frac{a+b+c}{2}\)
Mà a+b+c>2
\(\Rightarrow VT>1\) (đpcm)
Ta có: \(\dfrac{a^2}{b^2}+1\ge2\sqrt{\dfrac{a^2}{b^2}}=\dfrac{2a}{b}\)
Tương tự: \(\dfrac{b^2}{c^2}+1\ge\dfrac{2b}{c}\) ; \(\dfrac{c^2}{a^2}+1\ge\dfrac{2c}{a}\)
\(\Rightarrow\dfrac{a^2}{b^2}+\dfrac{b^2}{c^2}+\dfrac{c^2}{a^2}+3\ge\dfrac{2a}{b}+\dfrac{2b}{c}+\dfrac{2c}{a}\) (1)
Mà \(\dfrac{a}{b}+\dfrac{b}{c}+\dfrac{c}{a}\ge3\sqrt[3]{\dfrac{abc}{abc}}=3\)
\(\Rightarrow\dfrac{2a}{b}+\dfrac{2b}{c}+\dfrac{2c}{a}\ge\dfrac{a}{b}+\dfrac{b}{c}+\dfrac{c}{a}+3\) (2)
(1);(2) \(\Rightarrow\dfrac{a^2}{b^2}+\dfrac{b^2}{c^2}+\dfrac{c^2}{a^2}+3\ge\dfrac{a}{b}+\dfrac{b}{c}+\dfrac{c}{a}+3\)
\(\Rightarrow\dfrac{a^2}{b^2}+\dfrac{b^2}{c^2}+\dfrac{c^2}{a^2}\ge\dfrac{a}{b}+\dfrac{b}{c}+\dfrac{c}{a}\)
Dấu "=" xảy ra khi \(a=b=c\)
áp dụng BĐT: a^2/ x+ b^2/ y +c^2/ z >= (a+b+c)^2/ (x+y+z) cho BĐT trên
<=> a^2/(b+c) + b^2/ (a+c) + c/( a+b) >= (a+b+c)^2/ 2(a+b+c) =(a+b+c)/2
<=> a^2/(b+c) + b^2/ (a+c) + c/( a+b) >=(a+b+c)/2 (đpcm)
Dấu "=" xảy ra khi a=b=c