Ta có:
\(VT\ge\frac{1}{a^2b^2}+\frac{1}{b^2c^2}+\frac{1}{c^2a^2}=\frac{a}{a^3.b^2}+\frac{b}{b^3.c^2}+\frac{c}{c^3.a^2}\)
\(VT\ge\frac{a}{\frac{\left(a^3\right)^2+\left(b^2\right)^2}{2}}+\frac{b}{\frac{\left(b^3\right)^2+\left(c^2\right)^2}{2}}+\frac{c}{\frac{\left(c^3\right)^2+\left(a^2\right)^2}{2}}=VP\)
Dấu "=" xảy ra khi \(a=b=c=1\)
2. Cho a, b > 0. CM: \(\frac{1}{a}+\frac{1}{b}\ge\frac{4}{a+b}\)
Áp dụng CM các bđt sau:
a)Cho a, b, c > 0 thỏa mãn \(\frac{1}{a}+\frac{1}{b}+\frac{1}{c}=4.\) CM:\(\frac{1}{2a+b+c}+\frac{1}{a+2b+c}+\frac{1}{a+b+2c}\le1\)
b)\(\frac{ab}{a+b}+\frac{bc}{b+c}+\frac{ca}{c+a}\le\frac{a+b=c}{2}\left(a,b,c>0\right)\)
cho a,b,c,d >0 thỏa a+b+c+d=4 chứng minh \(\frac{a}{1+b^2c}+\frac{b}{1+c^2a}+\frac{c}{1+d^2a}+\frac{d}{1+a^2b}\)
cho a,b,c > 0. Cmr: \(\frac{ab}{a+3b+2c}+\frac{bc}{b+3c+2a}+\frac{ca}{c+3a+2b}\le\frac{a+b+c}{6}\)
cho \(c\ge b\ge a>0\) . Cmr: \(\frac{2a^2}{b+c}+\frac{2b^2}{c+a}+\frac{2c^2}{a+b}\le\frac{a^2}{b}+\frac{b^2}{c}+\frac{c^2}{a}\)
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 a,b,c> 0 . Cmr:
\(2\left(\frac{a}{b+2c}+\frac{b}{c+2a}+\frac{c}{a+2b}\right)\ge1+\frac{b}{b+2a}+\frac{c}{c+2b}+\frac{a}{a+2c}\)
cho a,b,c dương. Chứng minh \(\frac{1}{2b+c}+\frac{1}{2c+a}+\frac{1}{2a+b}\ge\frac{3}{a+b+c}\)
Cho a, b, c dương.
Cmr: \(\frac{1}{2a+b+c}+\frac{1}{a+2b+c}+\frac{1}{a+b+2c}\le\frac{1}{a+3b}+\frac{1}{b+3c}+\frac{1}{c+3a}\)
Cho a, b, c >0. Cm:
\(\frac{1}{a^4+1}+\frac{1}{1+b^4}+\frac{1}{1+c^4}\ge\frac{1}{1+ab^3}+\frac{1}{1+bc^3}+\frac{1}{1+ca^3}\)
Cho a,b,c là các số thực thỏa mãn \(a,b\ge-2\)và a+b+2c=6. CMR
a,\(a^2+b^2+4ab+16\ge 4c^2-16c+20\)
b,\(\frac{4-b^2}{4[(c-2)^2+1]}-\frac{a^2}{(a-b)^2+6ab+16}+5\ge0\)
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