Cho \(a,b.c>0\). Chứng minh rằng
\(\dfrac{a^3b}{c}+\dfrac{a^3c}{b}+\dfrac{b^3c}{a}+\dfrac{b^3a}{c}+\dfrac{c^3b}{a}+\dfrac{c^3a}{b}\ge6abc\).
Cho ba số a; b; c > 0 thỏa mãn: \(\dfrac{a+b-3c}{c}=\dfrac{b+c-3a}{a}=\dfrac{c+a-3b}{b}\)
Chứng minh rằng a = b =c.
Áp dụng tính chất dãy tỉ số bằng nhau ta có:
\(\dfrac{a+b-3c}{c}=\dfrac{b+c-3a}{a}=\dfrac{c+a-3b}{b}=\dfrac{a+b-3c+b+c-3a+c+a-3b}{c+a+b}=\dfrac{-\left(a+b+c\right)}{a+b+c}=-1\)
\(\dfrac{a+b-3c}{c}=-1\Rightarrow a+b-3c=-c\Rightarrow a+b-2c=0\left(1\right)\)
\(\dfrac{b+c-3a}{a}=-1\Rightarrow b+c-3a=-a\Rightarrow b+c-2a=0\left(2\right)\)
\(\dfrac{c+a-3b}{b}=-1\Rightarrow a+c-3b=-b\Rightarrow a+c-2b=0\left(3\right)\)
Từ (1), (2) ta có:\(a+b-2c=b+c-2a\Rightarrow3a=3c\Rightarrow a=c\left(4\right)\)
Từ (1), (3) ta có:\(a+b-2c=a+c-2b\Rightarrow3b=3c\Rightarrow b=c\left(5\right)\)
Từ (4), (5)\(\Rightarrow a=b=c\)
Cho a,b,c là ba số dương thỏa mãn a + b +c = 3 . Chứng minh rằng : \(\dfrac{\sqrt{3a+bc}}{a+\sqrt{3a+bc}}+\dfrac{\sqrt{3b+ac}}{b+\sqrt{3b+ac}}+\dfrac{\sqrt{3c+ab}}{c+\sqrt{3c+ab}}\) ≥ 2
Cho a,b,c là 3 số dương thỏa mãn a+b+c=3. Chứng minh rằng :\(\dfrac{\sqrt{3a+bc}}{a+\sqrt{3a+bc}}+\dfrac{\sqrt{3b+ac}}{b+\sqrt{3b+ac}}+\dfrac{\sqrt{3c+ab}}{c+\sqrt{3c+ab}}\)≥ 2
Cho \(\dfrac{a}{b}=\dfrac{c}{d}\). Chứng minh:
1) \(\dfrac{2a+3c}{2b+3d}=\dfrac{2a-3c}{2b-3d}\)
2) \(\dfrac{4a-3b}{4c-3d}=\dfrac{4a+3b}{4c+3d}\)
3) \(\dfrac{3a+5b}{3a-5b}=\dfrac{3c+5d}{3c-5d}\)
4) \(\dfrac{3a-7b}{b}=\dfrac{3c-7d}{d}\)
Đặt \(\dfrac{a}{b}=\dfrac{c}{d}=k\)
=>\(a=bk;c=dk\)
1: \(\dfrac{2a+3c}{2b+3d}=\dfrac{2\cdot bk+3\cdot dk}{2b+3d}=\dfrac{k\left(2b+3d\right)}{2b+3d}=k\)
\(\dfrac{2a-3c}{2b-3d}=\dfrac{2bk-3dk}{2b-3d}=\dfrac{k\left(2b-3d\right)}{2b-3d}=k\)
Do đó: \(\dfrac{2a+3c}{2b+3d}=\dfrac{2a-3c}{2b-3d}\)
2: \(\dfrac{4a-3b}{4c-3d}=\dfrac{4\cdot bk-3b}{4\cdot dk-3d}=\dfrac{b\left(4k-3\right)}{d\left(4k-3\right)}=\dfrac{b}{d}\)
\(\dfrac{4a+3b}{4c+3d}=\dfrac{4bk+3b}{4dk+3d}=\dfrac{b\left(4k+3\right)}{d\left(4k+3\right)}=\dfrac{b}{d}\)
Do đó: \(\dfrac{4a-3b}{4c-3d}=\dfrac{4a+3b}{4c+3d}\)
3: \(\dfrac{3a+5b}{3a-5b}=\dfrac{3bk+5b}{3bk-5b}=\dfrac{b\left(3k+5\right)}{b\left(3k-5\right)}=\dfrac{3k+5}{3k-5}\)
\(\dfrac{3c+5d}{3c-5d}=\dfrac{3dk+5d}{3dk-5d}=\dfrac{d\left(3k+5\right)}{d\left(3k-5\right)}=\dfrac{3k+5}{3k-5}\)
Do đó: \(\dfrac{3a+5b}{3a-5b}=\dfrac{3c+5d}{3c-5d}\)
4: \(\dfrac{3a-7b}{b}=\dfrac{3bk-7b}{b}=\dfrac{b\left(3k-7\right)}{b}=3k-7\)
\(\dfrac{3c-7d}{d}=\dfrac{3dk-7d}{d}=\dfrac{d\left(3k-7\right)}{d}=3k-7\)
Do đó: \(\dfrac{3a-7b}{b}=\dfrac{3c-7d}{d}\)
Xét a,b,c thỏa mãn a + b + c = 6. Chứng minh rằng \(\dfrac{a}{3a+1}+\dfrac{b}{3b+1}+\dfrac{c}{3c+1}\)≤ \(\dfrac{6}{7}\)
Lời giải:
$3\text{VT}=\frac{3a}{3a+1}+\frac{3b}{3b+1}+\frac{3c}{3c+1}$
$=1-\frac{1}{3a+1}+1-\frac{1}{3b+1}+1-\frac{1}{3c+1}$
$=3-\left[\frac{1}{3a+1}+\frac{1}{3b+1}+\frac{1}{3c+1}\right]$
Áp dụng BĐT Cauchy-Schwarz:
$\frac{1}{3a+1}+\frac{1}{3b+1}+\frac{1}{3c+1}\geq \frac{9}{3a+1+3b+1+3c+1}=\frac{9}{3(a+b+c)+3}=\frac{9}{3.6+3}=\frac{3}{7}$
$\Rightarrow 3\text{VT}\leq 3-\frac{3}{7}=\frac{18}{7}$
$\Rightarrow \text{VT}\leq \frac{6}{7}$ (đpcm)
Dấu "=" xảy ra khi $a=b=c=2$
Cho a, b, c > 0. Chứng minh \(\dfrac{a}{3a+b+c}+\dfrac{b}{3b+a+c}+\dfrac{c}{3c+a+b}\le\dfrac{3}{5}\)
Cho a,b,c là các số dương
CMR: \(6abc\le\dfrac{a^3b}{c}+\dfrac{b^3c}{a}+\dfrac{c^3a}{b}+\dfrac{a^3c}{b}+\dfrac{b^3a}{c}+\dfrac{c^3b}{a}\)
Áp dụng bđt AM-GM:
\(\dfrac{a^3b}{c}+\dfrac{b^3c}{a}+\dfrac{c^3a}{b}+\dfrac{a^3c}{b}+\dfrac{b^3a}{c}+\dfrac{c^3b}{a}\ge6\sqrt[6]{\dfrac{a^8b^8c^8}{a^2b^2c^2}}=6\sqrt[6]{a^6b^6c^6}=6abc\)Dấu "=" xảy ra khi \(a=b=c\)
Cho a,b,c là các số thực dương. Chứng minh rằng:
\(\dfrac{1}{a+2b+c}+\dfrac{1}{b+2c+a}+\dfrac{1}{c+2a+b}\le\dfrac{1}{a+3b}+\dfrac{1}{b+3c}+\dfrac{1}{c+3a}\)
Cho a,b,c>0 và dãy tỉ số\(\dfrac{2b+c-a}{a}=\dfrac{2c-b+a}{b}=\dfrac{2a+b-c}{c}\)
Tính P = \(\dfrac{\left(3a-2b\right)\left(3b-2c\right)\left(3c-2a\right)}{\left(3a-c\right)\left(3b-a\right)\left(3c-b\right)}\)
\(\dfrac{2b+c-a}{a}=\dfrac{2c-b+a}{b}=\dfrac{2a+b-c}{c}=\dfrac{2\left(a+b+c\right)}{a+b+c}=2\\ \Leftrightarrow\left\{{}\begin{matrix}2b+c-a=2a\\2c-b+a=2b\\2a+b-c=2c\end{matrix}\right.\Leftrightarrow\left\{{}\begin{matrix}3a-2b=c\\3b-2c=a\\3c-2a=b\end{matrix}\right.\text{ và }\left\{{}\begin{matrix}3a-c=2b\\3b-a=2c\\3c-b=2a\end{matrix}\right.\\ \Leftrightarrow P=\dfrac{a\cdot b\cdot c}{2a\cdot2b\cdot3c}=\dfrac{1}{8}\)