Cho 3 số thực dương a;b;c. Chứng minh:
\(\dfrac{2a^3}{a^6+bc}+\dfrac{2b^3}{b^6+ca}+\dfrac{2c^3}{c^6+ab}\le\dfrac{a}{bc}+\dfrac{b}{ca}+\dfrac{c}{ab}\)
Cho 3 số thực a,b,c .Chứng minh rằng :
\(\dfrac{2a^3}{a^6+bc}+\dfrac{2b^3}{b^6+ac}+\dfrac{2c^3}{c^6+ab}\le\dfrac{a}{bc}+\dfrac{b}{ac}+\dfrac{c}{ab}\)
Ta có: \(a^2+b^2+c^2\ge ab+bc+ca\ge\sqrt[]{abc}\left(\sqrt{a}+\sqrt{b}+\sqrt{c}\right)\)
Do đó:
\(VT\le\dfrac{2a^3}{2\sqrt{a^6bc}}+\dfrac{2b^3}{2\sqrt{b^6ac}}+\dfrac{2c^3}{2\sqrt{c^3ab}}=\dfrac{\sqrt{a}+\sqrt{b}+\sqrt{c}}{\sqrt{abc}}=\dfrac{\sqrt{abc}\left(\sqrt{a}+\sqrt{b}+\sqrt{c}\right)}{abc}\)
\(\le\dfrac{a^2+b^2+c^2}{abc}=\dfrac{a}{bc}+\dfrac{b}{ca}+\dfrac{c}{ab}\)
Dấu "=" xảy ra khi \(a=b=c=1\)
Cho các số thực dương a,b,c thỏa mãn a + b + c = 6. Chứng minh bất đẳng thức:
\(\dfrac{ab}{6+2b+c}+\dfrac{bc}{6+2c+a}+\dfrac{ca}{6+2a+b}\le1\).
Lời giải:
Vì \(a+b+c=6\) nên BĐT cần chứng minh tương đương với:
\(\frac{ab}{2b+c+a+b+c}+\frac{bc}{2c+a+a+b+c}+\frac{ca}{2a+b+a+b+c}\leq 1(*)\)
Thật vậy, áp dụng BĐT Cauchy-Schwarz ta có:
\(\frac{ab}{2b+c+a+b+c}=\frac{ab}{(b+c)+(c+a)+2b}\leq \frac{ab}{9}\left(\frac{1}{b+c}+\frac{1}{c+a}+\frac{1}{2b}\right)\)
Hoàn toàn tương tự:
\(\frac{bc}{2c+a+a+b+c}\leq \frac{bc}{9}\left(\frac{1}{a+b}+\frac{1}{a+c}+\frac{1}{2c}\right)\)
\(\frac{ca}{2a+b+a+b+c}\leq \frac{ca}{9}\left(\frac{1}{a+b}+\frac{1}{b+c}+\frac{1}{2a}\right)\)
Cộng các BĐT vừa thu được lại ta có:
\(\text{VT}\leq \frac{1}{9}\left(\frac{ab+ac}{b+c}+\frac{ab+bc}{a+c}+\frac{bc+ca}{a+b}+\frac{a+b+c}{2}\right)\)
\(\Leftrightarrow \text{VT}\leq \frac{1}{9}\left(a+b+c+\frac{a+b+c}{2}\right)=\frac{1}{9}\left(6+\frac{6}{2}\right)=1\)
BĐT \((*)\) hoàn tất, ta có đpcm.
Dấu bằng xảy ra khi \(a=b=c=2\)
Lời giải:
Áp dụng BĐT Cauchy-Schwarz ta có:
\(\text{VT}=\frac{ab}{6+2b+c}+\frac{bc}{6+2c+a}+\frac{ca}{6+2a+b}=\frac{ab}{a+b+c+2b+c}+\frac{bc}{a+b+c+2c+a}+\frac{ca}{a+b+c+2a+b}\)
\(=\frac{ab}{2b+(a+c)+(b+c)}+\frac{bc}{2c+(a+b)+(a+c)}+\frac{ca}{2a+(b+a)+(b+c)}\)
\(\leq \frac{ab}{9}\left(\frac{1}{2b}+\frac{1}{a+c}+\frac{1}{b+c}\right)+\frac{bc}{9}\left(\frac{1}{2c}+\frac{1}{a+b}+\frac{1}{a+c}\right)+\frac{ca}{9}\left(\frac{1}{2a}+\frac{1}{b+a}+\frac{1}{b+c}\right)\)
\(\text{VT}\leq \frac{a+b+c}{18}+\frac{ab+bc}{9(a+c)}+\frac{ab+ac}{9(b+c)}+\frac{bc+ac}{9(a+b)}\)
\(\text{VT}\leq \frac{(a+b+c)}{6}=\frac{6}{6}=1\) (đpcm)
Dấu "=" xảy ra khi $a=b=c=2$
Cho a, b, c là các số dương. Chứng minh rằng:
\(\dfrac{ab}{a+b+2c}+\dfrac{bc}{b+c+2a}+\dfrac{ca}{c+a+2b}\le\dfrac{a+b+c}{4}\)
\(\dfrac{bc}{a+b+c+a}\le\dfrac{bc}{4}\cdot\left(\dfrac{1}{a+b}+\dfrac{1}{a+c}\right)\\ \dfrac{ac}{b+c+a+b}\le\dfrac{ac}{4}\cdot\left(\dfrac{1}{b+c}+\dfrac{1}{a+b}\right)\\ \dfrac{ab}{a+c+b+c}\le\dfrac{ab}{4}\cdot\left(\dfrac{1}{a+c}+\dfrac{1}{b+c}\right)\\ \Leftrightarrow VT\le\dfrac{1}{a+b}\left(\dfrac{bc}{4}+\dfrac{ac}{4}\right)+\dfrac{1}{a+c}\left(\dfrac{bc}{4}+\dfrac{ab}{4}\right)+\dfrac{1}{b+c}\left(\dfrac{ac}{4}+\dfrac{ab}{4}\right)\\ =\dfrac{1}{a+b}\cdot\dfrac{c\left(a+b\right)}{4}+\dfrac{1}{a+c}\cdot\dfrac{b\left(a+c\right)}{4}+\dfrac{1}{b+c}\cdot\dfrac{a\left(b+c\right)}{4}\\ =\dfrac{c}{4}+\dfrac{b}{4}+\dfrac{a}{4}\\ =\dfrac{a+b+c}{4}\left(đfcm\right)\)
Cho các số thực dương a , b , c
Chứng minh rằng \(\dfrac{bc}{2a+b+c}+\dfrac{ca}{a+2b+c}+\dfrac{ab}{a+b+2c}\le\dfrac{a+b+c}{4}\)
Áp dụng bất đẳng thức \(\dfrac{1}{a+b}\le\dfrac{1}{4}\left(\dfrac{1}{a}+\dfrac{1}{b}\right)\) với a , b > 0
\(\Rightarrow\left\{{}\begin{matrix}\dfrac{bc}{2a+b+c}=\dfrac{bc}{a+b+a+c}\le\dfrac{bc}{4}\left(\dfrac{1}{a+b}+\dfrac{1}{a+c}\right)\\\dfrac{ca}{a+2b+c}=\dfrac{ca}{a+b+b+c}\le\dfrac{ca}{4}\left(\dfrac{1}{a+b}+\dfrac{1}{b+c}\right)\\\dfrac{ab}{a+b+2c}=\dfrac{ab}{a+c+b+c}\le\dfrac{ab}{4}\left(\dfrac{1}{a+c}+\dfrac{1}{b+c}\right)\end{matrix}\right.\)
\(\Rightarrow VT\le\dfrac{bc}{4}\left(\dfrac{1}{a+b}+\dfrac{1}{a+c}\right)+\dfrac{ca}{4}\left(\dfrac{1}{a+b}+\dfrac{1}{b+c}\right)+\dfrac{ab}{4}\left(\dfrac{1}{a+c}+\dfrac{1}{b+c}\right)\)
\(\Rightarrow VT\le\dfrac{bc}{4\left(a+b\right)}+\dfrac{bc}{4\left(a+c\right)}+\dfrac{ca}{4\left(a+b\right)}+\dfrac{ca}{4\left(b+c\right)}+\dfrac{ab}{4\left(a+c\right)}+\dfrac{ab}{4\left(b+c\right)}\)
\(\Rightarrow VT\le\left[\dfrac{bc}{4\left(a+b\right)}+\dfrac{ca}{4\left(a+b\right)}\right]+\left[\dfrac{bc}{4\left(a+c\right)}+\dfrac{ab}{4\left(a+c\right)}\right]+\left[\dfrac{ca}{4\left(b+c\right)}+\dfrac{ab}{4\left(b+c\right)}\right]\)
\(\Rightarrow VT\le\dfrac{bc+ca}{4\left(a+b\right)}+\dfrac{bc+ab}{4\left(a+c\right)}+\dfrac{ca+ab}{4\left(b+c\right)}\)
\(\Rightarrow VT\le\dfrac{c\left(a+b\right)}{4\left(a+b\right)}+\dfrac{b\left(c+a\right)}{4\left(a+c\right)}+\dfrac{a\left(b+c\right)}{4\left(b+c\right)}\)
\(\Rightarrow VT\le\dfrac{a+b+c}{4}\)
\(\Leftrightarrow\dfrac{bc}{2a+b+c}+\dfrac{ca}{a+2b+c}+\dfrac{ab}{a+b+2c}\le\dfrac{a+b+c}{4}\) ( đpcm )
Cho ba số thực dương a, b, c thoả mãn a+b+c=2 Chứng minh rằng:
\(\dfrac{ab}{\sqrt{2c+ab}}+\dfrac{bc}{\sqrt{2a+bc}}+\dfrac{ca}{\sqrt{2b+ca}}\le1\)
\(VT=\sqrt{\dfrac{a^2b^2}{c\left(a+b+c\right)+ab}}+\sqrt{\dfrac{b^2c^2}{a\left(a+b+c\right)+bc}}+\sqrt{\dfrac{a^2c^2}{b\left(a+b+c\right)+ac}}\\ VT=\sqrt{\dfrac{a^2b^2}{ac+ab+bc+c^2}}+\sqrt{\dfrac{b^2c^2}{a^2+ac+ab+bc}}+\sqrt{\dfrac{a^2c^2}{ab+bc+b^2+ac}}\\ VT=\sqrt{\dfrac{a^2b^2}{\left(c+a\right)\left(b+c\right)}}+\sqrt{\dfrac{a^2c^2}{\left(b+c\right)\left(a+b\right)}}+\sqrt{\dfrac{b^2c^2}{\left(a+b\right)\left(a+c\right)}}\)
Áp dụng BĐT Cauchy-Schwarz:
\(\Leftrightarrow\left\{{}\begin{matrix}\sqrt{\dfrac{b^2c^2}{\left(a+b\right)\left(a+c\right)}}\le\dfrac{\dfrac{bc}{a+b}+\dfrac{bc}{a+c}}{2}\\\sqrt{\dfrac{a^2c^2}{\left(a+b\right)\left(b+c\right)}}\le\dfrac{\dfrac{ca}{a+b}+\dfrac{ca}{b+c}}{2}\\\sqrt{\dfrac{a^2b^2}{\left(b+c\right)\left(a+c\right)}}\le\dfrac{\dfrac{ab}{b+c}+\dfrac{ab}{a+c}}{2}\end{matrix}\right.\)
\(\Rightarrow VT\le\dfrac{\left(\dfrac{bc}{a+b}+\dfrac{ca}{a+b}\right)+\left(\dfrac{ca}{b+c}+\dfrac{ab}{b+c}\right)+\left(\dfrac{bc}{a+c}+\dfrac{ab}{a+c}\right)}{2}\\ \Rightarrow VT\le\dfrac{a+b+c}{2}=\dfrac{2}{2}=1\)
Dấu \("="\Leftrightarrow a=b=c=\dfrac{2}{3}\)
cho a, b, c >0. chứng minh:
\(\dfrac{ab}{a+3b+2c}+\dfrac{bc}{b+3c+2a}+\dfrac{ca}{c+3a+2b}\le\dfrac{a+b+c}{6}\)
C/m BĐT : \(\dfrac{1}{x}+\dfrac{1}{y}+\dfrac{1}{z}\ge\dfrac{9}{x+y+z}\)
Áp dụng BĐT Sơ-vác-sơ:
\(\dfrac{1}{x}+\dfrac{1}{y}+\dfrac{1}{z}\ge\dfrac{\left(1+1+1\right)^2}{x+y+z}\ge\dfrac{9}{x+y+z}\)
Ta có: \(9\dfrac{ab}{a+3b+2c}=\dfrac{9ab}{\left(a+c\right)+\left(b+c\right)+2b}\le\dfrac{ab}{a+c}+\dfrac{ab}{b+c}+\dfrac{a}{2}\left(1\right)\)
CM tương tự
\(\dfrac{9bc}{b+3c+2a}\le\dfrac{bc}{a+c}+\dfrac{bc}{a+b}+\dfrac{b}{2}\left(2\right)\)
\(\dfrac{9ca}{c+3a+2b}\le\dfrac{ca}{b+c}+\dfrac{ca}{a+b}+\dfrac{c}{2}\left(3\right)\)
Cộng vế (1), (2), (3) => đpcm
Cho a, b, c > 0 . CMR :
\(\dfrac{ab}{a+3b+2c}+\dfrac{bc}{b+3c+2a}+\dfrac{ca}{c+3a+2b}\le\dfrac{a+b+c}{6}\)
\(\dfrac{ab}{a+3b+2c}=\dfrac{ab}{\left(a+c\right)+\left(b+c\right)+2b}\le\dfrac{ab}{9}\left(\dfrac{1}{a+c}+\dfrac{1}{b+c}+\dfrac{1}{2b}\right)=\dfrac{1}{9}\left(\dfrac{ab}{a+c}+\dfrac{ab}{b+c}+\dfrac{a}{2}\right)\)
Tương tự:
\(\dfrac{bc}{b+3c+2a}\le\dfrac{1}{9}\left(\dfrac{bc}{a+b}+\dfrac{bc}{c+a}+\dfrac{b}{2}\right)\)
\(\dfrac{ca}{c+3a+2b}\le\dfrac{1}{9}\left(\dfrac{ca}{b+c}+\dfrac{ca}{a+b}+\dfrac{c}{2}\right)\)
Cộng vế:
\(VT\le\dfrac{1}{9}\left(\dfrac{bc+ca}{a+b}+\dfrac{ca+ab}{b+c}+\dfrac{bc+ab}{c+a}+\dfrac{a+b+c}{2}\right)=\dfrac{a+b+c}{6}\)
Dấu "=" xảy ra khi \(a=b=c\)
Ch a, b, c là 3 số dương thỏa mãn: a+b+c=6. Tìm giá trị lớn nhất của biểu thức: \(A=\dfrac{ab}{a+3b+2c}+\dfrac{bc}{b+3c+2a}+\dfrac{ca}{c+3a+2b}\)
Áp dụng bđt \(\dfrac{9}{a+b+c}\le\dfrac{1}{a}+\dfrac{1}{b}+\dfrac{1}{c}\)
Khi đó \(\dfrac{9.ab}{a+3b+2c}=ab.\dfrac{9}{\left(a+c\right)+\left(c+b\right)+2b}\le\dfrac{ab}{a+c}+\dfrac{ab}{c+b}+\dfrac{a}{2}\)
Tương tự và cộng theo vế suy ra \(9A\le\dfrac{3\left(a+b+c\right)}{2}=9< =>A\le1\)
Dấu "=" xảy ra khi và chỉ khi a = b = c = 2
Cho các số thực dương \(a;b;c\) thỏa mãn :\(ab+bc+ca=abc\). Chứng minh rằng :
\(\dfrac{1}{a+2b+3c}+\dfrac{1}{b+2c+3a}+\dfrac{1}{c+2a+3b}\le\dfrac{1}{6}\).
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\(ab+bc+ca=abc\Rightarrow\dfrac{1}{a}+\dfrac{1}{b}+\dfrac{1}{c}=1\)
Đặt vế trái của BĐT cần chứng minh là P
Ta có:
\(\dfrac{1}{a+2b+3c}=\dfrac{1}{a+b+b+c+c+c}\le\dfrac{1}{6^2}\left(\dfrac{1}{a}+\dfrac{1}{b}+\dfrac{1}{b}+\dfrac{1}{c}+\dfrac{1}{c}+\dfrac{1}{c}\right)\)
\(\Rightarrow\dfrac{1}{a+2b+3c}\le\dfrac{1}{36}\left(\dfrac{1}{a}+\dfrac{2}{b}+\dfrac{3}{c}\right)\)
Tương tự:
\(\dfrac{1}{b+2c+3a}\le\dfrac{1}{36}\left(\dfrac{1}{b}+\dfrac{2}{c}+\dfrac{3}{a}\right)\) ; \(\dfrac{1}{c+2a+3b}\le\dfrac{1}{36}\left(\dfrac{1}{c}+\dfrac{2}{a}+\dfrac{3}{b}\right)\)
Cộng vế:
\(P\le\dfrac{1}{36}\left(\dfrac{6}{a}+\dfrac{6}{b}+\dfrac{6}{c}\right)=\dfrac{1}{6}\left(\dfrac{1}{a}+\dfrac{1}{b}+\dfrac{1}{c}\right)=\dfrac{1}{6}\)
Dấu "=" xảy ra khi \(a=b=c=\dfrac{1}{3}\)