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\(\frac{1}{a}+\frac{1}{b}+\frac{1}{c}=0\Leftrightarrow\frac{ab+bc+ca}{abc}=0\Rightarrow ab+bc+ca=0\\ \)
\(\Rightarrow bc=-ab-ac,ca=-ab-bc,ab=-bc-ca\)
\(\Rightarrow\frac{a^2+bc}{a^2+2bc}=\frac{a^2+bc}{a^2+bc+bc}=\frac{a^2+bc}{a^2+bc-ca-ab}=\frac{a^2+bc}{\left(a-b\right).\left(a-c\right)}\)
Làm tương tự. có: \(\frac{b^2+ca}{b^2+2ca}=\frac{b^2+ca}{b^2+ca-ab-bc}=\frac{b^2+ca}{\left(a-b\right).\left(c-b\right)}\)
\(\frac{c^2+ab}{c^2+2ab}=\frac{c^2+ab}{c^2+ab-ca-bc}=\frac{c^2+ab}{\left(b-c\right).\left(a-c\right)}\)
\(\Rightarrow A=\frac{a^2+bc}{\left(a-b\right).\left(a-c\right)}+\frac{b^2+ca}{\left(a-b\right).\left(c-b\right)}+\frac{c^2+ab}{\left(b-c\right).\left(a-c\right)}\)
\(=\frac{\left(a^2+bc\right).\left(b-c\right)}{\left(a-b\right).\left(b-c\right).\left(a-c\right)}-\frac{\left(b^2+ca\right).\left(a-c\right)}{\left(a-b\right).\left(b-c\right).\left(a-c\right)}+\frac{\left(c^2+ab\right).\left(a-b\right)}{\left(a-b\right).\left(b-c\right).\left(a-c\right)}\)
Sau đó bạn thực hiện tiếp nhé.
Bài 1: Cho \(a,b,c\ge0:a^2+b^2+c^2=3\). CMR: \(a^4b^4+b^4c^4+c^4a^4\le3\)
Bài 2: Cho \(a,b,c\ge0\). CMR: \(a^2+b^2+c^2+2abc+1\ge2\left(ab+bc+ca\right)\)
Bài 3: Cho \(a,b,c\ge0:a^2+b^2+c^2=a+b+c\). CMR: \(a^2b^2+b^2c^2+c^2a^2\le ab+bc+ca\)
Bài 4: Cho \(a,b,c\ge0\). CMR: \(4\left(a+b+c\right)^3\ge27\left(ab^2+bc^2+ca^2+abc\right)\)
Bài 5: Cho \(a,b,c\ge0:a+b+c=3\).CMR: \(\frac{1}{2bc^2+1}+\frac{1}{2ca^2+1}+\frac{1}{2ab^2+1}\ge1\)
Lời giải:
Xét tử :
\(\frac{a^2}{a^2+2bc}+\frac{b^2}{b^2+2ac}+\frac{c^2}{c^2+2ab}=\frac{a^2}{a^2+bc+(-ab-ac)}+\frac{b^2}{b^2+ac+(-ab-bc)}+\frac{c^2}{c^2+ab+(-bc-ac)}\)
\(=\frac{a^2}{a(a-b)-c(a-b)}+\frac{b^2}{b(b-c)-a(b-c)}+\frac{c^2}{c(c-a)-b(c-a)}\)
\(=\frac{a^2}{(a-c)(a-b)}+\frac{b^2}{(b-a)(b-c)}+\frac{c^2}{(c-a)(c-b)}\)
\(=\frac{a^2(c-b)+b^2(a-c)+c^2(b-a)}{(a-b)(b-c)(c-a)}\)
\(=\frac{(ab^2+bc^2+ca^2)-(a^2b+b^2c+c^2a)}{(ab^2+bc^2+ca^2)-(a^2b+b^2c+c^2a)}=1\)
Xét mẫu (tương tự bên tử)
\(\frac{bc}{a^2+2bc}+\frac{ac}{b^2+2ac}+\frac{ab}{c^2+2ab}=\frac{bc}{(a-c)(a-b)}+\frac{ac}{(b-a)(b-c)}+\frac{ab}{(c-a)(c-b)}\)
\(=\frac{bc(c-b)+ac(a-c)+ab(b-a)}{(a-b)(b-c)(c-a)}=\frac{(ab^2+bc^2+ca^2)-(a^2b+b^2c+c^2a)}{(a-b)(b-c)(c-a)}\)
\(=\frac{(ab^2+bc^2+ca^2)-(a^2b+b^2c+c^2a)}{(ab^2+bc^2+ca^2)-(a^2b+b^2c+c^2a)}=1\)
Do đó:
\(A=\frac{1}{1}=1\)
cố tử thần ♡๖ۣۜŦεαм♡❤Ɠ长♡ღ
Chị ơi dùng bđt BCS , dấu = xảy ra P =1 như thế có gọi là giá trị của P=1 không nhỉ ?
\(a^2+b^2+c^2=\left(a+b+c\right)^2\)
\(\Leftrightarrow a^2+b^2+c^2+2ab+2ac+2bc=a^2+b^2+c^2\)
\(\Leftrightarrow2\left(ab+ac+bc\right)=0\)
\(\Leftrightarrow ab+ac+bc=0\)
\(\Leftrightarrow\hept{\begin{cases}ab=-ac-bc\\ac=-ab-bc\\bc=-ab-ac\end{cases}}\)
Ta có : \(a^2+2bc=a^2+bc+bc=a^2+bc-ab-ac=a\left(a-b\right)-c\left(a-b\right)=\left(a-b\right)\left(a-c\right)\)
CMTT ta có : \(\hept{\begin{cases}b^2+2ac=\left(b-a\right)\left(b-c\right)\\c^2+2ab=\left(c-a\right)\left(c-b\right)\end{cases}}\)
Thay vào A ta được :
\(A=\frac{1}{\left(a-b\right)\left(a-c\right)}+\frac{1}{\left(b-a\right)\left(b-c\right)}+\frac{1}{\left(c-a\right)\left(c-b\right)}\)
\(A=\frac{b-c}{\left(a-b\right)\left(a-c\right)\left(b-c\right)}+\frac{-a+c}{\left(a-b\right)\left(a-c\right)\left(b-c\right)}+\frac{a-b}{\left(a-b\right)\left(a-c\right)\left(b-c\right)}\)
\(A=\frac{b-c-a+c+a-b}{\left(a-b\right)\left(a-c\right)\left(b-c\right)}\)
\(A=\frac{0}{\left(a-b\right)\left(a-c\right)\left(b-c\right)}\)
\(A=0\)
Ta có: \(\left(a+b+c\right)^2=a^2+b^2+c^2\)
\(\Leftrightarrow a^2+b^2+c^2+2\left(ab+bc+ca\right)=a^2+b^2+c^2\)
\(\Leftrightarrow ab+bc+ca=0\)
\(\Rightarrow\hept{\begin{cases}ab=-bc-ca\\bc=-ca-ab\\ca=-ab-bc\end{cases}}\)
Thay vào ta được: \(\frac{a^2}{a^2+2bc}=\frac{a^2}{a^2+bc-ca-ab}=\frac{a^2}{\left(a-b\right)\left(a-c\right)}\)
Tương tự: \(\frac{b^2}{b^2+2ca}=\frac{b^2}{\left(b-a\right)\left(b-c\right)}\) ; \(\frac{c^2}{c^2+2ab}=\frac{c^2}{\left(c-a\right)\left(c-b\right)}\)
\(\Rightarrow P=-\left[\frac{a^2}{\left(a-b\right)\left(c-a\right)}+\frac{b^2}{\left(b-c\right)\left(a-b\right)}+\frac{c^2}{\left(c-a\right)\left(b-c\right)}\right]\)
\(=-\left[\frac{a^2\left(b-c\right)+b^2\left(c-a\right)+c^2\left(a-b\right)}{\left(a-b\right)\left(b-c\right)\left(c-a\right)}\right]\)
\(=\frac{\left(b-c\right)\left(a^2+bc-ca-ab\right)}{\left(a-b\right)\left(b-c\right)\left(a-c\right)}\)
\(=\frac{\left(b-c\right)\left(a-b\right)\left(a-c\right)}{\left(a-b\right)\left(b-c\right)\left(a-c\right)}=1\)
\(\left(a+b+c\right)^2=a^2+b^2+c^2\Leftrightarrow ab+ac+bc=0\)
\(\frac{a^2}{a^2+2bc}=\frac{a^2}{a^2-ab-ac+bc}=\frac{a^2}{\left(a-b\right)\left(a-c\right)}\)
Tương tự: \(\frac{b^2}{b^2+2ac}=\frac{b^2}{\left(b-a\right)\left(b-c\right)};\frac{c^2}{c^2+2ac}=\frac{c^2}{\left(c-a\right)\left(c-b\right)}\)
\(P=\frac{a^2}{a^2+2bc}+\frac{b^2}{b^2+2ac}+\frac{c^2}{c^2+2ab}\)
\(=\frac{a^2}{\left(a-b\right)\left(a-c\right)}-\frac{b^2}{\left(a-b\right)\left(b-c\right)}+\frac{c^2}{\left(a-c\right)\left(b-c\right)}\)\(=\frac{\left(a-b\right)\left(a-c\right)\left(b-c\right)}{\left(a-b\right)\left(a-c\right)\left(b-c\right)}=1\)
a - b = 4 ; b - c = -2 (1)
=> a - b + b - c = 4 - 2
=> a - c = 2 (2)
(1) => a - b - b + c = 4 + 2
=> a - 2b + c = 6 (3)
\(T=\frac{a^2+b^2+c^2-ab-bc-ac}{a^2-c^2-2ab+2bc}\)
\(2T=\frac{2a^2+2b^2-2ab-2bc-2ac}{\left(a-c\right)\left(a+c\right)-2b\left(a-c\right)}\)
\(2T=\frac{\left(a-b\right)^2+\left(b-c\right)^2+\left(c-a\right)^2}{\left(a-c\right)\left(a+c-2b\right)}\) và (1)(2)(3)
\(\Rightarrow2T=\frac{4^2+\left(-2\right)^2+2^2}{2\cdot6}=2\)
\(\Rightarrow T=1\)