Cho a,b,c là độ dài 3 cạnh của 1 tam giác
a, CMR: ab(a+b-2c) + bc(b+c-2a) + ac(a+c-2b) \(\ge\) 0
b, CMR: \(\dfrac{a}{b+c-a}+\dfrac{b}{a+c-b}+\dfrac{c}{a+b-c}\ge3\)
Cho a,b,c là độ dài 3 cạnh của tam giác.
a, CMR:\(ab\left(a+b-2c\right)+bc\left(b+c-2a\right)+ac\left(a+c-2b\right)\ge0\)
b, CMR: \(\dfrac{a}{b+c-a}+\dfrac{b}{a+c-b}+\dfrac{c}{a+b-c}\ge3\)
Lời giải:
a)
Theo bất đẳng thức AM-GM ta có:
\(ab(a+b)+bc(b+c)+ac(c+a)\)
\(=a^2b+ab^2+b^2c+bc^2+c^2a+ca^2\geq 6\sqrt[6]{a^2b.ab^2.b^2c.bc^2.c^2a.ca^2}\)
\(\Leftrightarrow ab(a+b)+bc(b+c)+ca(c+a)\geq 6abc\)
\(\Leftrightarrow ab(a+b-2c)+bc(b+c-2a)+ca(c+a-2b)\geq 0\)
Ta có đpcm.
Dấu bằng xảy ra khi \(a=b=c\)
b) Áp dụng BĐT Cauchy-Schwarz:
\(\text{VT}=\frac{a^2}{ab+ac-a^2}+\frac{b^2}{ab+bc-b^2}+\frac{c^2}{ca+cb-c^2}\)
\(\geq \frac{(a+b+c)^2}{ab+ac-a^2+ab+bc-b^2+ca+cb-c^2}\)
\(\Leftrightarrow \text{VT}\geq \frac{(a+b+c)^2}{2(ab+bc+ac)-(a^2+b^2+c^2)}\)
Vì $a,b,c$ là độ dài ba cạnh tam giác nên
\(a(b+c-a)+b(a+c-b)+c(a+b-c)>0\)
hay \(2(ab+bc+ac)-(a^2+b^2+c^2)>0\)
Mặt khác theo BĐT AM-GM ta có:
\(a^2+b^2+c^2\geq ab+bc+ac\Rightarrow 2(ab+bc+ac)-(a^2+b^2+c^2)\leq ab+bc+ac\)
\(\Rightarrow \text{VT}\geq \frac{(a+b+c)^2}{ab+bc+ac}=\frac{a^2+b^2+c^2+2(ab+bc+ac)}{ab+bc+ac}\geq \frac{3(ab+bc+ac)}{ab+bc+ac}=3\)
Vậy ta có đpcm.
Dấu bằng xảy ra khi \(a=b=c\)
a, a,b,c>0. CMR:\(\dfrac{ab}{a+b+2c}+\dfrac{bc}{b+c+2a}+\dfrac{ac}{a+c+2b}\le\dfrac{a+b+c}{4}\)
b, a,b,c>0. CMR:\(\dfrac{ab}{a+3b+2c}+\dfrac{bc}{b+3c+2a}+\dfrac{ac}{c+3a+2b}\le\dfrac{a+b+c}{6}\)
a.
\(\sum\dfrac{ab}{a+c+b+c}\le\dfrac{1}{4}\sum\left(\dfrac{ab}{a+c}+\dfrac{ab}{b+c}\right)=\dfrac{a+b+c}{4}\)
2.
\(\dfrac{ab}{a+3b+2c}=\dfrac{ab}{a+b+2c+2b}\le\dfrac{ab}{9}\left(\dfrac{4}{a+b+2c}+\dfrac{1}{2b}\right)=4.\dfrac{ab}{a+b+2c}+\dfrac{a}{18}\)
Quay lại câu a
\(b,\dfrac{ab}{a+3b+2c}=\left(\dfrac{1}{9}ab\right)\cdot\dfrac{9}{\left(a+c\right)+\left(b+c\right)+2b}\le\left(\dfrac{1}{9}ab\right)\cdot\left(\dfrac{1}{a+c}+\dfrac{1}{b+c}+\dfrac{1}{2b}\right)=\dfrac{1}{9}\cdot\left(\dfrac{ab}{a+b}+\dfrac{ab}{b+c}+\dfrac{a}{2}\right)\)
Cmtt: \(\dfrac{bc}{b+3c+2a}\le\dfrac{1}{9}\cdot\left(\dfrac{bc}{a+b}+\dfrac{bc}{a+b}+\dfrac{b}{2}\right);\dfrac{ca}{c+3a+2b}\le\dfrac{1}{9}\cdot\left(\dfrac{ca}{b+c}+\dfrac{ca}{a+b}+\dfrac{c}{2}\right)\)
\(\Rightarrow VT\le\dfrac{1}{9}\left(\dfrac{bc+ca}{a+b}+\dfrac{ab+ac}{b+c}+\dfrac{ab+bc}{a+c}+\dfrac{a+b+c}{2}\right)\\ \le\dfrac{1}{9}\left(a+b+c+\dfrac{a+b+c}{2}\right)=\dfrac{1}{9}\cdot\dfrac{3}{2}\left(a+b+c\right)=\dfrac{a+b+c}{6}\)
Dấu $"="$ khi $a=b=c$
Cho a;b;c>0 tm a+b+c=3
CMR \(\dfrac{2b+c}{a}+\dfrac{2c+a}{b}+\dfrac{2a+b}{c}+\dfrac{18abc}{ab+bc+ac}\ge12\)
Cho \(a,b,c>0\) thỏa mãn \(ab+bc+ca=3\) . CMR : \(\sqrt[3]{\dfrac{a}{b\left(b+2c\right)}}+\sqrt[3]{\dfrac{b}{c\left(c+2a\right)}}+\sqrt[3]{\dfrac{c}{a\left(a+2b\right)}\ge\dfrac{3}{\sqrt[3]{3}}}\)
1. Cho a,b,c là độ dài 3 cạnh của 1 tam giác vuông, cạnh huyền là a. Cmr: a3 > b3 + c3
2. Cho a,b,c > 0 và a+b+c=4. CMR: ab/a+b+2c + bc/2a+b+c + ac/a+2b+c <= 1
Bài 1: Cho a,b,c là những số dương thỏa mãn: a+b+c=3
CMR: \(\dfrac{a^2}{a+2b^3}+\dfrac{b^2}{b+2c^3}+\dfrac{c^2}{c+2a^3}\ge1\)
Bài 2: Cho a, b, c thỏa mãn: ab+bc+ca=3
CMR: \(\dfrac{a}{2b^3+1}+\dfrac{b}{2c^3+1}+\dfrac{c}{2a^3+1}\ge1\)
Bài 3: Cho a, b, c > 0. CMR: \(\dfrac{a^2}{b}+\dfrac{b^2}{c}+\dfrac{c^2}{a}\ge a+3b\)
Dấu = xảy ra khi a=b=2c
Cho a, b, c là độ dài 3 cạnh của 1 tam giác. CMR: \(\dfrac{1}{a+b-c}+\dfrac{1}{b+c-a}+\dfrac{1}{c+a-b}\ge\dfrac{1}{a}+\dfrac{1}{b}+\dfrac{1}{c}\)
\(A=\dfrac{1}{a+b-c}+\dfrac{1}{b+c-a}+\dfrac{1}{c+a-b}\)\(\Rightarrow\left\{{}\begin{matrix}\dfrac{1}{a+b-c}+\dfrac{1}{b+c-a}\ge\dfrac{4}{a+b-c+b+c-a}\ge\dfrac{4}{2b}\ge\dfrac{2}{b}\\\dfrac{1}{b+c-a}+\dfrac{1}{c+a-b}\ge\dfrac{4}{b+c-a+c+a-b}\ge\dfrac{4}{2c}\ge\dfrac{2}{c}\\\dfrac{1}{a+b-c}+\dfrac{1}{c+a-b}\ge\dfrac{4}{a+b-c+c+a-b}\ge\dfrac{4}{2a}\ge\dfrac{2}{a}\end{matrix}\right.\)
\(\Rightarrow2\left(\dfrac{1}{a+b-c}+\dfrac{1}{b+c-a}+\dfrac{1}{c+a-b}\right)\ge\left(\dfrac{2}{a}+\dfrac{2}{b}+\dfrac{2}{c}\right)\ge2\left(\dfrac{1}{a}+\dfrac{1}{b}+\dfrac{1}{c}\right)\)
\(\Rightarrow A\ge\dfrac{1}{a}+\dfrac{1}{b}+\dfrac{1}{c}\) \(dấu"="xảy\) \(ra\Leftrightarrow a=b=c\)
Cho a, b, c>0 và a+b+c\(\ge3\)
Cmr:
\(\dfrac{a^2}{a+\sqrt{bc}}+\dfrac{b^2}{b+\sqrt{ac}}+\dfrac{c^2}{c+\sqrt{ab}}\ge\dfrac{3}{2}\)
Áp dụng bđt cosi schwart ta có:
`VT>=(a+b+c)^2/(a+b+c+sqrt{ab}+sqrt{bc}+sqrt{ca})`
Dễ thấy `sqrt{ab}+sqrt{bc}+sqrt{ca}<a+b+c`
`=>VT>=(a+b+c)^2/(2(a+b+c))=(a+b+c)/2=3`
Dấu "=" `<=>a=b=c=1.`
1. Cho a,b,c là độ dài 3 cạnh của 1 tam giác vuông, cạnh huyền là a. Cmr:
a3 > b3 + c3
2. Cho a,b,c > 0 và a+b+c=4. CMR
ab/a+b+2c + bc/2a+b+c + ac/a+2b+c <= 1