Cho \(a,b,c>0\).Chứng minh \(\frac{a^2}{2a+3b}+\frac{b^2}{2b+3c}+\frac{c^2}{2c+3a}\ge\frac{a+b+c}{5}\)
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.\(\frac{a^2}{2a+3b}+\frac{b^2}{2b+3c}+\frac{c^2}{2c+3a}\ge\frac{a+b+c}{5}\)
Áp dụng bđt Cauchy-schwarz dạng engel ta có:
1. \(\frac{a^2}{a+2b}+\frac{b^2}{b+2c}+\frac{c^2}{c+2a}\ge\frac{\left(a+b+c\right)^2}{\left(a+2b\right)+\left(b+2c\right)+\left(c+2a\right)}=\frac{a+b+c}{3}\)
Dấu "=" \(\Leftrightarrow\frac{a}{a+2b}=\frac{b}{b+2c}=\frac{c}{c+2a}\Leftrightarrow a=b=c\)
2. \(\frac{a^2}{2a+3b}+\frac{b^2}{2b+3c}+\frac{c^2}{2c+3a}\ge\frac{\left(a+b+c\right)^2}{\left(2a+3b\right)+\left(2b+3c\right)+\left(2c+3a\right)}=\frac{a+b+c}{5}\)
Dấu "=" \(\Leftrightarrow a=b=c\)
CMR: Với mọi a;b;c>0
\(\frac{2b+3c}{a+2b+3c}+\frac{2c+3a}{b+2c+3a}+\frac{2a+3b}{c+2a+3b}\ge\frac{5}{2}\)
cho a,b,c >0 chứng minh
\(\frac{a}{2b+3c}+\frac{b}{2c+3a}+\frac{c}{2a+3b}\ge\frac{3}{5}\)
\(P=\frac{a^2}{2ab+3ac}+\frac{b^2}{2bc+3ab}+\frac{c^2}{2ac+3bc}\)
\(P\ge\frac{\left(a+b+c\right)^2}{5\left(ab+bc+ca\right)}\ge\frac{3\left(ab+bc+ca\right)}{5\left(ab+bc+ca\right)}=\frac{3}{5}\)
Dấu "=" xảy ra khi \(a=b=c\)
Cho a,b,c là độ dài ba cạnh của một tam giác. Chứng minh rằng:
\(\frac{a}{2b+3c}+\frac{b}{2c+3a}+\frac{c}{3b+2a}\ge\frac{3}{5}\)
Ta có:
A = \(\frac{a}{2b+3c}+\frac{b}{2c+3a}+\frac{c}{3b+2a}=\frac{a^2}{2ab+3ac}+\frac{b^2}{2bc+3ab}+\frac{c^2}{3bc+2ac}\)
A \(\ge\frac{\left(a+b+c\right)^2}{2ab+3ac+2bc+3ab+3bc+2ac}\)(bđt svacxo \(\frac{x_1^2}{y_1}+\frac{x_2^2}{y_2}+\frac{x_3^2}{y_3}\ge\frac{\left(x_1+x_2+x_3\right)^2}{y_1+y_2+y_3}\))
A \(\ge\frac{\left(a+b+c\right)^2}{5\left(ab+bc+ac\right)}\ge\frac{\left(a+b+c\right)^2}{\frac{5\left(a+b+c\right)^2}{3}}\) (bđt \(xy+yz+xz\le\frac{\left(x+y+z\right)^2}{3}\)(*)
CM bđt * <=> \(3xy+3yz+3xz\le x^2+y^2+z^2+2xz+2xy+2yz\)
<=> \(\left(x-y\right)^2+\left(x-z\right)^2+\left(y-z\right)^2\ge0\) (luôn đúng)
<=> A \(\ge\frac{3}{5}\) --> ĐPCM
a) Cho a,b,c>0. chứng minh rằng:\(\frac{a}{3a^2+2b^2+c^2}+\frac{b}{3b^2+2c^2+a^2}+\frac{c}{3c^2+2a^2+b^2}\le\frac{1}{6}\left(\frac{1}{a}+\frac{1}{b}+\frac{1}{c}\right)\)
Cho a, b, c dương. Chứng minh: \(\frac{1}{a\sqrt{3a+2b}}+\frac{1}{b\sqrt{3b+2c}}+\frac{1}{c\sqrt{3c+2a}}\ge\frac{3}{\sqrt{5abc}}\)
Lời giải:
BĐT cần chứng minh tương đương với:
\(\frac{bc}{\sqrt{5abc(3a+2b)}}+\frac{ac}{\sqrt{5abc(3b+2c)}}+\frac{ab}{\sqrt{5abc(3c+2a)}}\geq \frac{3}{5}(*)\)
Áp dụng BĐT AM-GM:
\(5abc(3a+2b)=5ab.(3ac+2bc)\leq \left(\frac{5ab+3ac+2bc}{2}\right)^2\)
\(\Rightarrow \frac{bc}{\sqrt{5abc(3a+2b)}}\geq \frac{2bc}{5ab+3ac+2bc}=\frac{2(bc)^2}{5ab^2c+3abc^2+2b^2c^2}\)
Hoàn toàn tương tự với các phân thức còn lại, cộng theo vế ta suy ra:
\(\sum \frac{bc}{\sqrt{5abc(3a+2b)}}\geq \sum \frac{2(bc)^2}{5ab^2c+3abc^2+2b^2c^2}(1)\)
Áp dụng BĐT Cauchy_Schwarz và AM-GM:
\(\sum \frac{2(bc)^2}{5ab^2c+3abc^2+2b^2c^2}\geq 2.\frac{(bc+ab+ac)^2}{2[(ab)^2+(bc)^2+(ca)^2+4abc(a+b+c)]}=\frac{(ab+bc+ac)^2}{(ab)^2+(bc)^2+(ca)^2+4abc(a+b+c)}\)
\(=\frac{(ab+bc+ac)^2}{(ab+bc+ac)^2+2abc(a+b+c)}\geq \frac{(ab+bc+ac)^2}{(ab+bc+ac)^2+\frac{2}{3}(ab+bc+ac)^2}=\frac{3}{5}(2)\)
Từ $(1);(2)$ suy ra $(*)$ đúng. BĐT được chứng minh.
Dấu "=" xảy ra khi $a=b=c$
Cho a, b, c, d là các số thực dương. Chứng minh :
\(\frac{a}{b+2c+3d}+\frac{b}{c+2d+3a}+\frac{c}{d+2a+3b}+\frac{d}{a+2b+3c}\ge\frac{2}{3}\)
Cho \(a;b;c;d>0\)chứng minh
\(\frac{a}{b+2c+3d}+\frac{b}{c+2d+3a}+\frac{c}{d+2a+3b}+\frac{d}{a+2b+3c}\ge\frac{2}{3}\)
cre: dự vào đề tóan quốc tế mỹ
\(\text{Σ}\frac{a}{b+2c+3d}=\text{Σ}\frac{a^2}{ab+2ac+3ad}\ge\frac{\left(a+b+c+d\right)^2}{6\left(ab+bc+cd+ad\right)}\)
\(=\frac{\left(a+b\right)^2+\left(c+d\right)^2+2\left(a+b\right)\left(c+d\right)}{6\left(ab+bc+cd+ad\right)}=\frac{a^2+c^2+b^2+d^2+2ab+2cd+2\left(a+b\right)\left(c+d\right)}{6\left(ab+bc+cd+ad\right)}\)
\(\ge\frac{4\left(ab+bc+cd+ad\right)}{6\left(ab+bc+cd+ad\right)}=\frac{2}{3}\)
Dấu = xảy ra khi a=b=c=d
Áp dụng BĐT Cauchy-Schwarz ta có:
\(\frac{a}{b+2c+3d}+\frac{b}{c+2d+3a}+\frac{c}{d+2a+3b}+\frac{d}{a+2b+3c}\)
\(=\frac{a^2}{ab+2ac+3ad}+\frac{b^2}{bc+2bd+3ab}+\frac{c^2}{cd+2ac+3bc}+\frac{d^2}{ad+2bd+3cd}\)
\(\ge\frac{\left(a+b+c+d\right)^2}{4.\left(ab+ad+bc+bd+ca+cd\right)}\)\(\ge\frac{\left(a+b+c+d\right)^2}{\frac{3}{2}.\left(a+b+c+d\right)^2}=\frac{2}{3}\)
Dấu "=" xảy ra \(\Leftrightarrow a=b=c=d\)
\(VT=\frac{a^2}{ab+2ac+3ad}+\frac{b^2}{bc+2bd+3ab}+\frac{c^2}{cd+2ac+3bc}+\frac{d^2}{ad+2bd+3cd}\)
Áp dụng BĐT Svac-xơ cho 3 số dương ta được :
\(VT\ge\frac{\left(a+b+c+d\right)^2}{4ab+4ac+4ad+4bc+4bd+4cd}\)
Áp dụng BĐT phụ \(x^2+y^2\ge2xy\) ta được :
\(a^2+b^2\ge2ab;a^2+c^2\ge2ac;a^2+d^2\ge2ad\)
\(b^2+c^2\ge2bc;b^2+d^2\ge2bd;c^2+d^2\ge2cd\)
\(\Rightarrow3\left(a^2+b^2+c^2+d^2\right)\ge2\left(ab+ac+ad+bc+bd+cd\right)\)
Ta lại có : \(\left(a+b+c+d\right)^4=a^2+b^2+c^2+d^2+2ab+2ac+2ad+2bc+2bd+2cd\)
\(\ge\frac{8\left(ab+ac+ad+bc+bd+cd\right)}{3}\)
\(\Rightarrow VT\ge\frac{\left(a+b+c+d\right)^4}{4\left(ab+ac+ad+bc+bd+cd\right)}\ge\frac{8\left(ab+ac+ad+bc+bd+cd\right)}{12\left(ab+ac+ad+bc+bd+cd\right)}=\frac{2}{3}\)
Dấu "=" xảy ra khi \(a=b=c=d\)
Khó quá!
Cho \(a,b,c>0\). Chứng minh rằng:
\(\frac{a^4}{3a^3+2b^3}+\frac{b^4}{3b^3+2c^3}+\frac{c^4}{3c^3+2a^3}\ge\frac{a+b+c}{5}\)