1/ a/dung bđt Cauchy - Schwarz dạng phân thức: \(\frac{a^2}{b+3c}+\frac{b^2}{c+3a}+\frac{c^2}{a+3b}\ge\frac{\left(a+b+c\right)^2}{4\left(a+b+c\right)}=\frac{a+b+c}{4}=\frac{3}{4}\)
2/ a/dung bđt bunhiacopxki :
\(S^2=\left(\sqrt{a+b}+\sqrt{b+c}+\sqrt{c+a}\right)^2\le\left(1^2+1^2+1^2\right)\left(a+b+b+c+c+a\right)=3\cdot2\left(a+b+c\right)=6\cdot6=36\)
=> \(S\le6\)
C/m BĐT : \(\frac{5b^3-a^3}{ab+3b^2}+\frac{5c^3-b^3}{bc+3c^2}+\frac{5a^3-c^3}{ca+3a^2}\le a+b+c\)
\(\frac{c+a}{\sqrt{a^2+c^2}}\ge\frac{c+b}{\sqrt{c^2+b^2}};a>b>0,c>\sqrt{ab}\)
cho a b c > 0. Chứng minh các bất đẳng thức :
1, \(\left(a+b+c\right)\left(\frac{1}{a}+\frac{1}{b}+\frac{1}{c}\right)\ge9\)
2, \(\frac{1}{a}+\frac{1}{b}+\frac{1}{c}+\frac{1}{d}\ge\frac{16}{a+b+c+d}\)
3, ( 1+a+b) (a+b+ab) \(\ge9ab\)
4, \(\left(\sqrt{a}+\sqrt{b}\right)^8\ge64ab\left(a+b\right)^2\)
5, \(3a^3+7b^3\ge9ab^2\)
6, \(\left(\sqrt{a}+\sqrt{b}\right)^2\ge2\sqrt{2\left(a+b\right)\sqrt{ab}}\)
cho các số a,b,c > 0. chứng minh:
1.\(\frac{a^2}{a+2b}+\frac{b^2}{b+2c}+\frac{c^2}{c+2a}\ge\frac{a+b+c}{3}\)
2.\(\frac{a^2}{2a+3b}+\frac{b^2}{2b+3c}+\frac{c^2}{2c+3a}\ge\frac{a+b+c}{5}\)
C/m các BĐT sau :
\(1.a^3-3a+4\ge b^3-3b
\)
\(2,\frac{1}{\frac{1}{a+c}+\frac{1}{b+d}}\ge\frac{1}{\frac{1}{a}+\frac{1}{b}}+\frac{1}{\frac{1}{c}+\frac{1}{d}}\) với a, b, c, d>0
\(3,a^3+b^3\ge\frac{1}{4};a+b\ge1\)
4, \(a^3+b^3\le a^4+b^4;a+b\ge2\)
5, \(\left(a+b\right)\left(a^3+b^3\right)\left(a^5+b^5\right)\le4\left(a^9+b^9\right);a,b\ge0\)
6, \(\frac{c+a}{\sqrt{a^2+c^2}}\ge\frac{c+b}{\sqrt{c^2+b^2}};a>b>0,c>\sqrt{ab}\)
Các bn làm đc bài nào thì giúp mk với, cảm ơn ạ !
Cho các số thực dương a,b, c thỏa mãn a+b+c=3. Chứng minh rằng:
\(\sqrt{a^3+3b}\) + \(\sqrt{b^3+3c}\) + \(\sqrt{c^3+3a}\) ≥ 6
1/cho số a >0 tìm GTNN của P = 2a +\(\frac{4}{a}\)+\(\frac{16}{a+2}\)
2/ cho a,b,c là số thực ϵ [0;\(\frac{1}{4}\)) chứng minh:
\(\sqrt{a\left(1-4a\right)}+\sqrt{b\left(1-4b\right)}+\sqrt{c\left(1-4c\right)}\le\frac{3}{4}\)
3/ cho các số dương a,b,c tỏa abc = 1. Chứng minh
\(\frac{1}{a^2c+b^2c+1}+\frac{1}{b^2a+c^2a+1}+\frac{1}{c^2b+a^2b+1}\le1\)
cho a , b , c >0. Chứng minh các bất đẳng thức :
1, ab + bc + ca \(\ge\sqrt{abc}\left(\sqrt{a}+\sqrt{b}+\sqrt{c}\right)\)
2, \(\frac{ab}{c}+\frac{bc}{a}+\frac{ac}{b}\ge a+b+c\)
3, \(ab+\frac{a}{b}+\frac{b}{a}\ge a+b+1\)
4, \(\frac{a^3}{b}+\frac{b^3}{c}+\frac{c^3}{a}\ge ab+bc+ca\)
5, \(\frac{a}{bc}+\frac{b}{ca}+\frac{c}{ab}\ge\frac{1}{a}+\frac{1}{b}+\frac{1}{c}\)
Cho a, b, c là các số thực dương. Chứng minh rằng :
\(\frac{a}{\sqrt{4a^2+\left(b+c\right)^2}}+\frac{b}{\sqrt{4b^2+\left(c+a\right)^2}}+\frac{c}{\sqrt{4c^2+\left(a+b\right)^2}}\le\frac{3\sqrt{2}}{4}\)
cho a,b,c>0. chứng minh rằng:
\(\sqrt{\frac{\left(a^2+bc\right)\left(b+c\right)}{a\left(b^2+c^2\right)}}\) +\(\sqrt{\frac{\left(b^2+ac\right)\left(a+c\right)}{b\left(a^2+c^2\right)}}\) +\(\sqrt{\frac{\left(c^2+ab\right)\left(a+b\right)}{c\left(a^2+b^2\right)}}\) \(\ge\) \(3\sqrt{2}\)
