![](https://rs.olm.vn/images/avt/0.png?1311)
Hãy nhập câu hỏi của bạn vào đây, nếu là tài khoản VIP, bạn sẽ được ưu tiên trả lời.
![](https://rs.olm.vn/images/avt/0.png?1311)
![](https://rs.olm.vn/images/avt/0.png?1311)
Từ giả thiết:
\(a^2+b^2+c^2+a^2+b^2+c^2+2\left(ab+bc+ca\right)\le4\)
\(\Rightarrow a^2+b^2+c^2+ab+bc+ca\le2\)
Ta có:
\(\dfrac{ab+1}{\left(a+b\right)^2}=\dfrac{1}{2}.\dfrac{2ab+2}{\left(a+b\right)^2}\ge\dfrac{1}{2}.\dfrac{2ab+a^2+b^2+c^2+ab+bc+ca}{\left(a+b\right)^2}=\dfrac{1}{2}\dfrac{\left(a+b\right)^2+\left(a+c\right)\left(b+c\right)}{\left(a+b\right)^2}\)
\(=\dfrac{1}{2}+\dfrac{1}{2}.\dfrac{\left(a+c\right)\left(b+c\right)}{\left(a+b\right)^2}\)
Tương tự và cộng lại, đồng thời đặt \(\left(a+b;b+c;c+a\right)=\left(x;y;z\right)\):
\(\Rightarrow VT\ge\dfrac{3}{2}+\dfrac{1}{2}\left(\dfrac{yz}{x^2}+\dfrac{xz}{y^2}+\dfrac{xy}{z^2}\right)\ge\dfrac{3}{2}+\dfrac{1}{2}.3\sqrt[3]{\dfrac{yz.xz.xy}{x^2y^2z^2}}=3\) (đpcm)
Dấu "=" xảy ra khi \(a=b=c=\dfrac{1}{\sqrt{3}}\)
![](https://rs.olm.vn/images/avt/0.png?1311)
\(\Leftrightarrow\dfrac{2a}{b+c}+\dfrac{2b}{c+a}+\dfrac{2c}{a+b}\ge3+\dfrac{2a^2+2b^2+2c^2-2\left(ab+bc+ca\right)}{\left(a+b+c\right)^2}\)
\(\Leftrightarrow\dfrac{2a}{b+c}+\dfrac{2b}{c+a}+\dfrac{2c}{a+b}\ge5-\dfrac{6\left(ab+bc+ca\right)}{\left(a+b+c\right)^2}\)
\(\Leftrightarrow\dfrac{2a}{b+c}+\dfrac{2b}{c+a}+\dfrac{2c}{a+b}+\dfrac{6\left(ab+bc+ca\right)}{\left(a+b+c\right)^2}\ge5\)
Do \(\dfrac{2a}{b+c}+\dfrac{2b}{c+a}+\dfrac{2c}{a+b}=\dfrac{2a^2}{ab+ac}+\dfrac{2b^2}{bc+ab}+\dfrac{2c^2}{ac+bc}\ge\dfrac{\left(a+b+c\right)^2}{ab+bc+ca}\)
Nên ta chỉ cần chứng minh:
\(\dfrac{\left(a+b+c\right)^2}{ab+bc+ca}+\dfrac{6\left(ab+bc+ca\right)}{\left(a+b+c\right)^2}\ge5\)
Điều này hiển nhiên đúng do:
\(VT=\dfrac{2}{3}.\dfrac{\left(a+b+c\right)^2}{ab+bc+ca}+\dfrac{6\left(ab+bc+ca\right)}{\left(a+b+c\right)^2}+\dfrac{\left(a+b+c\right)^2}{3\left(ab+bc+ca\right)}\)
\(VT\ge2\sqrt{\dfrac{12\left(a+b+c\right)^2\left(ab+bc+ca\right)}{3\left(ab+bc+ca\right)\left(a+b+c\right)^2}}+\dfrac{3\left(ab+bc+ca\right)}{3\left(ab+bc+ca\right)}=5\)
Dấu "=" xảy ra khi \(a=b=c\)
![](https://rs.olm.vn/images/avt/0.png?1311)
Áp dụng bất đẳng thức Cauchy - Schwarz
\(\Rightarrow VT\ge3\sqrt[3]{\left[\left(1+\dfrac{1}{a}\right)\left(1+\dfrac{1}{b}\right)\left(1+\dfrac{1}{c}\right)\right]^4}\)
\(\Rightarrow VT\ge3\left(\sqrt[3]{1+\dfrac{1}{a}+\dfrac{1}{b}+\dfrac{1}{c}+\dfrac{1}{ab}+\dfrac{1}{bc}+\dfrac{1}{ca}+\dfrac{1}{abc}}\right)^4\) (1)
Áp dụng bất đẳng thức Cauchy - Schwarz
\(\Rightarrow\left\{{}\begin{matrix}\dfrac{1}{a}+\dfrac{1}{b}+\dfrac{1}{c}\ge3\sqrt[3]{\dfrac{1}{abc}}\\\dfrac{1}{ab}+\dfrac{1}{bc}+\dfrac{1}{ca}\ge3\sqrt[3]{\dfrac{1}{a^2b^2c^2}}\end{matrix}\right.\)
\(\Rightarrow1+\dfrac{1}{a}+\dfrac{1}{b}+\dfrac{1}{c}+\dfrac{1}{ab}+\dfrac{1}{bc}+\dfrac{1}{ca}+\dfrac{1}{abc}\ge1+3\sqrt[3]{\dfrac{1}{abc}}+3\sqrt[3]{\dfrac{1}{a^2b^2c^2}}+\dfrac{1}{abc}\)
\(\Rightarrow1+\dfrac{1}{a}+\dfrac{1}{b}+\dfrac{1}{c}+\dfrac{1}{ab}+\dfrac{1}{bc}+\dfrac{1}{ca}+\dfrac{1}{abc}\ge\left(1+\dfrac{1}{\sqrt[3]{abc}}\right)^3\)
\(\Rightarrow3\left(\sqrt[3]{1+\dfrac{1}{a}+\dfrac{1}{b}+\dfrac{1}{c}+\dfrac{1}{ab}+\dfrac{1}{bc}+\dfrac{1}{ca}+\dfrac{1}{abc}}\right)^4\ge3\left(1+\dfrac{1}{\sqrt[3]{abc}}\right)^4\) (2)
Áp dụng bất đẳng thức Cauchy - Schwarz
\(\Rightarrow\sqrt[3]{abc}\le\dfrac{abc+1+1}{3}=\dfrac{abc+2}{3}\)
\(\Rightarrow1+\dfrac{1}{\sqrt[3]{abc}}\ge1+\dfrac{3}{abc+2}\)
\(\Rightarrow3\left(1+\dfrac{1}{\sqrt[3]{abc}}\right)^4\ge3\left(1+\dfrac{3}{abc+2}\right)^4\) (3)
Từ (1) và (2) và (3)
\(\Rightarrow VT\ge3\left(1+\dfrac{3}{abc+2}\right)^4\)
\(\Leftrightarrow\left(1+\dfrac{1}{a}\right)^4+\left(1+\dfrac{1}{b}\right)^4+\left(1+\dfrac{1}{c}\right)^4\ge3\left(1+\dfrac{3}{abc+2}\right)^4\) ( đpcm )
![](https://rs.olm.vn/images/avt/0.png?1311)
Áp dụng BĐT holder cho n bộ 3 số:
\(\left(\sum\dfrac{b^nc^n}{b+c}\right)\left[\sum\left(b+c\right)\right]\left(1+1+1\right)..\left(1+1+1\right)\ge\left(ab+bc+ca\right)^n\)
\(\Leftrightarrow VT\ge\dfrac{\left(ab+bc+ca\right)^n}{3^{n-2}.2.\left(a+b+c\right)}\ge\dfrac{3^{n-2}.3abc\left(a+b+c\right)}{3^{n-2}.2.\left(a+b+c\right)}=\dfrac{3}{2}\)
#Hint:(\(\left\{{}\begin{matrix}ab+bc+ca\ge3\\\left(ab+bc+ca\right)^2\ge3abc\left(a+b+c\right)\end{matrix}\right.\))
BĐT holder thường dùng:
\(\left(a_1^m+a_2^m+...+a_k^m\right)\left(b_1^m+b_2^m+...+b_k^m\right)...\left(c_1^m+...+c_k^m\right)\ge\left(a_1b_1...c_1+a_2.b_2...c_2+...+a_k.b_k...c_k\right)^m\)
trong đó VT có m thừa số từ a đến c
![](https://rs.olm.vn/images/avt/0.png?1311)
Áp dụng BĐT \(\dfrac{1}{x}+\dfrac{1}{y}+\dfrac{1}{z}\ge\dfrac{9}{x+y+z}\) ta được
\(\dfrac{1}{2a}+\dfrac{1}{2b}+\dfrac{1}{2b}\ge\dfrac{9}{2\left(a+2b\right)}\)
\(\dfrac{1}{2b}+\dfrac{1}{2c}+\dfrac{1}{2c}\ge\dfrac{9}{2\left(b+2c\right)}\)
\(\dfrac{1}{2c}+\dfrac{1}{2a}+\dfrac{1}{2a}\ge\dfrac{9}{2\left(c+2a\right)}\)
Cộng các BĐT theo vế
\(\dfrac{3}{2}\left(\dfrac{1}{a}+\dfrac{1}{b}+\dfrac{1}{c}\right)\ge\dfrac{9}{2}\left(\dfrac{1}{a+2b}+\dfrac{1}{b+2c}+\dfrac{1}{c+2a}\right)\)
\(\Rightarrow\dfrac{1}{a}+\dfrac{1}{b}+\dfrac{1}{c}\ge3\left(\dfrac{1}{a+2b}+\dfrac{1}{b+2c}+\dfrac{1}{c+2a}\right)\)
Dấu " = " xảy ra khi a = b = c ( a,b,c > 0 )
![](https://rs.olm.vn/images/avt/0.png?1311)
Đề sai rồi: a,b,c > 0 thì làm sao mà có: ab + bc + ca = 0 được.
![](https://rs.olm.vn/images/avt/0.png?1311)
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 \(a=b=c\)
\(b\left(a-b\right)\le\dfrac{\left(b+a-b\right)^2}{4}=\dfrac{a^2}{4}\)
\(\Rightarrow\dfrac{1}{b\left(a-b\right)}\ge\dfrac{4}{a^2}\)
\(\Rightarrow a+\dfrac{1}{b\left(a-b\right)}\ge a+\dfrac{4}{a^2}=\dfrac{a}{2}+\dfrac{a}{2}+\dfrac{4}{a^2}\ge3\sqrt[3]{\dfrac{a}{2}\dfrac{a}{2}\dfrac{4}{a^2}}=3\)
Dấu "=" xảy ra khi \(\left\{{}\begin{matrix}\dfrac{a}{2}=\dfrac{4}{a^2}\\b=a-b\end{matrix}\right.\) \(\Rightarrow\left\{{}\begin{matrix}a=2\\b=1\end{matrix}\right.\)