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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.
\(\sum\dfrac{ab}{a+c+b+c}\le\dfrac{1}{4}\sum\left(\dfrac{ab}{a+c}+\dfrac{ab}{b+c}\right)=\dfrac{a+b+c}{4}\)
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
\(A=\dfrac{1}{a+b-c}+\dfrac{1}{b+c-a}+\dfrac{1}{c+a-b}\)\(\Rightarrow\left\{{}\begin{matrix}\dfrac{1}{a+b-c}+\dfrac{1}{b+c-a}\ge\dfrac{4}{a+b-c+b+c-a}\ge\dfrac{4}{2b}\ge\dfrac{2}{b}\\\dfrac{1}{b+c-a}+\dfrac{1}{c+a-b}\ge\dfrac{4}{b+c-a+c+a-b}\ge\dfrac{4}{2c}\ge\dfrac{2}{c}\\\dfrac{1}{a+b-c}+\dfrac{1}{c+a-b}\ge\dfrac{4}{a+b-c+c+a-b}\ge\dfrac{4}{2a}\ge\dfrac{2}{a}\end{matrix}\right.\)
\(\Rightarrow2\left(\dfrac{1}{a+b-c}+\dfrac{1}{b+c-a}+\dfrac{1}{c+a-b}\right)\ge\left(\dfrac{2}{a}+\dfrac{2}{b}+\dfrac{2}{c}\right)\ge2\left(\dfrac{1}{a}+\dfrac{1}{b}+\dfrac{1}{c}\right)\)
\(\Rightarrow A\ge\dfrac{1}{a}+\dfrac{1}{b}+\dfrac{1}{c}\) \(dấu"="xảy\) \(ra\Leftrightarrow a=b=c\)
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.`
Ta có bđt \(\left(x+y+z\right)\left(\dfrac{1}{x}+\dfrac{1}{y}+\dfrac{1}{z}\right)\ge9\)(1)
Chứng minh:
Áp dụng bđt cosi cho 3 số dương:
\(x+y+z\ge3\sqrt[3]{xyz}\left(2\right)\)
\(\dfrac{1}{x}+\dfrac{1}{y}+\dfrac{1}{z}\ge3\sqrt[3]{\dfrac{1}{xyz}}\)(3)
Từ (2),(3)\(\Rightarrow\left(x+y+z\right)\left(\dfrac{1}{x}+\dfrac{1}{y}+\dfrac{1}{z}\right)\ge3\sqrt[3]{xyz}.3\sqrt[3]{\dfrac{1}{xyz}}=9\)
Vậy bđt (1) đã chứng minh
Áp dụng bđt (1), ta có \(\left[\left(2a+b\right)+\left(2b+c\right)+\left(2c+a\right)\right]\left(\dfrac{1}{2a+b}+\dfrac{1}{2b+c}+\dfrac{1}{2c+a}\right)\ge9\Leftrightarrow3\left(a+b+c\right)\left(\dfrac{1}{2a+b}+\dfrac{1}{2b+c}+\dfrac{1}{2c+a}\right)\ge9\Leftrightarrow3.1.\left(\dfrac{1}{2a+b}+\dfrac{1}{2b+c}+\dfrac{1}{2c+a}\right)\ge9\Leftrightarrow\dfrac{1}{2a+b}+\dfrac{1}{2b+c}+\dfrac{1}{2c+a}\ge3\)Vậy nếu a+b+c=1 thì \(\dfrac{1}{2a+b}+\dfrac{1}{2b+c}+\dfrac{1}{2c+a}\ge3\)
\(\dfrac{ab}{a+3b+2c}=\dfrac{ab}{\left(a+c\right)+\left(b+c\right)+2b}\le\dfrac{ab}{9}\left(\dfrac{1}{a+c}+\dfrac{1}{b+c}+\dfrac{1}{2b}\right)=\dfrac{1}{9}\left(\dfrac{ab}{a+c}+\dfrac{ab}{b+c}+\dfrac{a}{2}\right)\)
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 \(a=b=c\)
a)Bunhia:
\(\left(1+2\right)\left(b^2+2a^2\right)\ge\left(1.b+\sqrt{2}.\sqrt{2}a\right)^2=\left(b+2a\right)^2\)
b)\(ab+bc+ca=abc\Leftrightarrow\dfrac{1}{a}+\dfrac{1}{b}+\dfrac{1}{c}=1\)
Áp dụng bđt câu a
=>VT\(\ge\)\(\dfrac{b+2a}{\sqrt{3}ab}+\dfrac{c+2b}{\sqrt{3}bc}+\dfrac{a+2c}{\sqrt{3}ca}\)
\(\Leftrightarrow VT\ge\dfrac{1}{a}+\dfrac{2}{b}+\dfrac{1}{b}+\dfrac{2}{c}+\dfrac{1}{c}+\dfrac{2}{a}=3=VP\)
Tự tìm dấu "="
Nguyễn Việt LâmMashiro ShiinaBNguyễn Thanh HằngonkingCẩm MịcFa CTRẦN MINH HOÀNGhâu DehQuân Tạ MinhTrương Thị Hải Anh
