Áp dụng bđt Bunhia-cốp-xki ở dạng phân thức, ta có:
\(\dfrac{1}{a+b}+\dfrac{1}{b+c}+\dfrac{1}{c+a}\ge\dfrac{\left(1+1+1\right)^2}{a+b+b+c+c+a}=\dfrac{9}{2}\)( vì a+b+c=1)
Dấu bằng xảy ra \(\Leftrightarrow\dfrac{1}{a+b}=\dfrac{1}{b+c}=\dfrac{1}{c+a}\Leftrightarrow a+b=b+c=c+a\Leftrightarrow a=b=c=\dfrac{1}{3}\)(vì a+b+c=1)
Cho a,b,c > 0 và a+b+c ≤ 1. CMR: A = \(\frac{1}{a^2+2bc}+\frac{1}{b^2+2ca}+\frac{1}{c^2+2ab}\) ≥ 9
Bài 1: Cho a,b,c \(\ge\)0. CMR: \(\frac{b+c}{a}+\frac{c+a}{b}+\frac{a+b}{c}\ge6\)
Bài 2: Cho a,b,c \(\ge\)0. CMR: \(\frac{a}{b+c}+\frac{b}{c+a}+\frac{c}{a+b}\ge\frac{3}{2}\)
(4)Bài 1:Với \(\forall\) a>b>0. CMR: a+ \(\frac{1}{b\left(a-b\right)}\ge3\)
(7) Bài 2: Cho a,b,c \(\ne\) 0 .CMR: \(\frac{a^2}{b^2}+\frac{b^2}{c^2}+\frac{c^2}{a^2}\ge\frac{b}{a}+\frac{c}{b}+\frac{a}{c}\)
(8) Bài 3: Cho a,b,c>0 thõa mãn abc=1
CMR: \(\frac{b+c}{\sqrt{a}}+\frac{c+a}{\sqrt{b}}+\frac{a+b}{\sqrt{c}}\ge\sqrt{a}+\sqrt{b}+\sqrt{c}+3\)
Cho a, b, c>0 và a+b+c=abc. CMR: \(\frac{b}{a^2}+\frac{c}{b^2}+\frac{a}{c^2}+3\ge\left(\frac{1}{a}+\frac{1}{b}+\frac{1}{c}\right)^2+\sqrt{3}\)
1. Cho a,b,c>0. Cmr: \(\frac{a}{\left(b+c\right)^2}+\frac{b}{\left(c+a\right)^2}+\frac{c}{\left(a+b\right)^2}\ge\frac{9}{4\left(a+b+c\right)}\)
Cho a,b,c>0 thoả mãn : a+b+c=3. CMR: \(\frac{1}{a^2+1}+\frac{1}{b^2+1}+\frac{1}{c^2+1}\ge\frac{3}{2}\)
Cho a+b+c = 3 ;a,b,c > 0
CMR : \(\frac{a^{ }}{1+b^2}+\frac{b}{1+c^2}+\frac{c}{1+a^2}\)≥\(\frac{3}{2}\)
Cho a,b,c>0 CMR \(\frac{a}{b}+\frac{b}{c}+\frac{c}{a}\ge\frac{a+b}{b+c}+\frac{b+c}{a+b}+1\)
cho \(\left\{{}\begin{matrix}a,b,c>0\\a+b+c=abc\end{matrix}\right.\).CMR: \(\frac{b}{a^2}+\frac{c}{b^2}+\frac{a}{c^2}+3\ge\left(\frac{1}{a}+\frac{1}{b}+\frac{1}{c}\right)^2+\sqrt{3}\)
