\(VT\ge4\frac{\sqrt[4]{bc}}{\sqrt{a}}.4\frac{\sqrt[4]{ca}}{\sqrt{b}}.4\frac{\sqrt[4]{ab}}{\sqrt{c}}=64\)
\(VT\ge4\frac{\sqrt[4]{bc}}{\sqrt{a}}.4\frac{\sqrt[4]{ca}}{\sqrt{b}}.4\frac{\sqrt[4]{ab}}{\sqrt{c}}=64\)
Cho ba số thực dương a, b, c. Chứng minh rằng:
a) \(\left(a+\frac{4b}{c^2}\right)\left(b+\frac{4c}{a^2}\right)\left(c+\frac{4a}{b^2}\right)\ge64\)
b) \(\frac{a^3}{b}+\frac{b^3}{c}+\frac{c^3}{a}\ge ab+bc+ca\)
1) \(\frac{a}{b+c}+\frac{b}{c+a}+\frac{c}{a+b}\ge\frac{3}{2}\)
2) với \(\left\{{}\begin{matrix}a,b,c>0\\a+b+c=3\end{matrix}\right.\) chứng minh \(\frac{a^3}{b\left(2c+a\right)}+\frac{b^3}{c\left(2a+b\right)}+\frac{c^3}{a\left(2b+c\right)}\ge1\)
cho cac so thuc duong a b c thoa a^2+b^2+c^2>=3 chung minh
\(\frac{\left(a+1\right)\left(b+2\right)}{\left(b+1\right)\left(b+5\right)}+\frac{\left(b+1\right)\left(c+2\right)}{\left(c+1\right)\left(c+5\right)}+\frac{\left(c+1\right)\left(a+2\right)}{\left(a+1\right)\left(a+5\right)}\ge\frac{3}{2}\)
Cho a,b,c >0 abc=1. CMR \(\frac{a^4}{b^2\left(c+a\right)}+\frac{b^4}{c^2\left(a+b\right)}+\frac{c^4}{a^2\left(b+c\right)}\ge\frac{a+b+c}{2}\)
Cho c\(\ge\)b\(\ge\)a>0. Chứng minh \(b\left(\frac{1}{a}+\frac{1}{c}\right)+\frac{1}{b}\left(a+c\right)\le\left(a+c\right)\left(\frac{1}{a}+\frac{1}{c}\right)\)
Cho a,b,c>0.CMR:
\(\frac{8\left(a^2+b^2+c^2\right)}{ab+bc+ca}+\frac{27\left(a+b\right)\left(b+c\right)\left(c+a\right)}{\left(a+b+c\right)^3}\ge16\)
Help!
1. Cho a > b > 0 .Chứng minh rằng :
\(a,a+\frac{1}{b\left(a-b\right)}\ge3\)
\(b,a+\frac{4}{\left(a-b\right)\left(b+1\right)^2}\ge3\)
\(c,a+\frac{1}{b\left(a-b\right)^2}\ge2\sqrt{2}\)
Mọi người ơi giúp mình với
Câu 1: Cho x, y, z > 0 và \(5\left(x^2+y^2+z^2\right)=6\left(xy+yz+xz\right)\)Tìm giá trị nhỏ nhất của
\(P=\left(x+y+z\right)\left(\frac{1}{x}+\frac{1}{y}+\frac{1}{z}\right)\)
Câu 2: Cho a, b, c >0 và \(\left\{{}\begin{matrix}ab+bc+ca>0\\a\ge c\end{matrix}\right.\)Tìm giá trị nhỏ nhất của
\(p=\frac{\left(a+b\right)}{\left(b+c\right)}+\frac{\left(b+c\right)}{\left(c+a\right)}+\frac{\left(c+a\right)^2}{a\left(b+c\right)+c\left(b+a\right)}\)
cho a,b c đôi một khác nhau. Cmr:
\(\frac{\left(a+b\right)^2}{\left(a-b\right)^2}+\frac{\left(b+c\right)^2}{\left(b-c\right)^2}+\frac{\left(c+a\right)^2}{\left(c-a\right)^2}\ge2\)