cho \(a,b,c\ge0\),\(a+b+c=4.\)TIM GTLN P=\(\sqrt{a^3b+b^3c+c^3a+abc^2}+\sqrt{ab^3+bc^3+ca^3+bca^2}\)
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9 tháng 8 2020
gt <=> \(\frac{1}{ab}+\frac{1}{bc}+\frac{1}{ca}=1\)
Đặt: \(\frac{1}{a}=x;\frac{1}{b}=y;\frac{1}{c}=z\)
=> Thay vào thì \(VT=\frac{\frac{1}{xy}}{\frac{1}{z}\left(1+\frac{1}{xy}\right)}+\frac{1}{\frac{yz}{\frac{1}{x}\left(1+\frac{1}{yz}\right)}}+\frac{1}{\frac{zx}{\frac{1}{y}\left(1+\frac{1}{zx}\right)}}\)
\(VT=\frac{z}{xy+1}+\frac{x}{yz+1}+\frac{y}{zx+1}=\frac{x^2}{xyz+x}+\frac{y^2}{xyz+y}+\frac{z^2}{xyz+z}\ge\frac{\left(x+y+z\right)^2}{x+y+z+3xyz}\)
Có BĐT x, y, z > 0 thì \(\left(x+y+z\right)\left(xy+yz+zx\right)\ge9xyz\)Ta thay \(xy+yz+zx=1\)vào
=> \(x+y+z\ge9xyz=>\frac{x+y+z}{3}\ge3xyz\)
=> Từ đây thì \(VT\ge\frac{\left(x+y+z\right)^2}{x+y+z+\frac{x+y+z}{3}}=\frac{3}{4}\left(x+y+z\right)\ge\frac{3}{4}.\sqrt{3\left(xy+yz+zx\right)}=\frac{3}{4}.\sqrt{3}=\frac{3\sqrt{3}}{4}\)
=> Ta có ĐPCM . "=" xảy ra <=> x=y=z <=> \(a=b=c=\sqrt{3}\)
AH
Akai Haruma
Giáo viên
8 tháng 2 2018
Lời giải:
Áp dụng BĐT Bunhiacopxky:
\((3a^2+b^2)(3+1)\geq (3a+b)^2\Rightarrow \sqrt{3a^2+b^2}\ge \frac{3a+b}{2}\)
\(\Rightarrow \frac{ab}{\sqrt{3a^2+b^2}+1}\leq \frac{2ab}{3a+b+2}\)
Thực hiện tương tự với các phân thức còn lại và cộng theo vế:
\(\Rightarrow Q\leq \frac{2ab}{3a+b+2}+\frac{2bc}{3b+c+2}+\frac{2ac}{3c+a+2}\)
\(\Leftrightarrow 3Q\leq \frac{6ab}{3a+b+2}+\frac{6bc}{3b+c+2}+\frac{6ac}{3c+a+2}\)
\(\Leftrightarrow 3Q\le 2b-\frac{2b^2+4b}{3a+b+2}+2c-\frac{2c^2+4c}{3b+c+2}+2a-\frac{2a^2+4a}{3c+a+2}\)
\(\Leftrightarrow 3Q\leq 6-\left(\frac{2b^2+4b}{3a+b+2}+\frac{2c^2+4c}{3b+c+2}+\frac{2a^2+4a}{3c+a+2}\right)(1)\)
Áp dụng BĐT Cauchy-Schwarz:
\(\frac{2b^2}{3a+b+2}+\frac{2c^2}{3b+c+2}+\frac{2a^2}{3c+a+2}\geq \frac{2(b+c+a)^2}{3a+b+2+3b+c+2+3c+a+2}=\frac{2(a+b+c)^2}{4(a+b+c)+6}=1(2)\)
Và:
\(\frac{4b}{3a+b+2}+\frac{4c}{3b+c+2}+\frac{4a}{3c+a+2}=4\left(\frac{b^2}{3ab+b^2+2b}+\frac{c^2}{3bc+c^2+2c}+\frac{a^2}{3ac+a^2+2a}\right)\)
\(\geq \frac{4(b+c+a)^2}{3ab+b^2+2b+3bc+c^2+3ac+a^2+2a}=\frac{4(a+b+c)^2}{(a+b+c)^2+2(a+b+c)+(ab+bc+ac)}\)
\(\geq \frac{4(a+b+c)^2}{(a+b+c)^2+2(a+b+c)+\frac{(a+b+c)^2}{3}}=2(3)\) (AM-GM)
Từ \((1); (2); (3)\Rightarrow 3Q\leq 6-(2+1)\Leftrightarrow 3Q\leq 3\Leftrightarrow Q\leq 1\)
Vậy Q(max) là $1$
Dấu bằng xảy ra khi \(a=b=c=1\)
20 tháng 5 2018
Ta có: \(P=\dfrac{bc}{\sqrt{3a+bc}}+\dfrac{ca}{\sqrt{3b+ca}}+\dfrac{ab}{\sqrt{3c+ab}}\)
\(=\dfrac{bc}{\sqrt{\left(a+b+c\right)a+bc}}+\dfrac{ca}{\sqrt{\left(a+b+c\right)b+ca}}+\dfrac{ab}{\sqrt{\left(a+b+c\right)+ab}}\)\(=\dfrac{bc}{\sqrt{a^2+ab+ac+bc}}+\dfrac{ca}{\sqrt{ab+b^2+bc+ca}}+\dfrac{ab}{\sqrt{c^2+ac+ab+bc}}\)\(=\dfrac{bc}{\sqrt{\left(a+b\right)\left(a+c\right)}}+\dfrac{ca}{\sqrt{\left(b+c\right)\left(b+a\right)}}+\dfrac{ab}{\sqrt{\left(c+a\right)\left(c+b\right)}}\le\)\(\le\dfrac{1}{2}\left(\dfrac{b^2}{a+b}+\dfrac{c^2}{a+c}+\dfrac{c^2}{b+c}+\dfrac{a^2}{a+b}+\dfrac{a^2}{a+c}+\dfrac{b^2}{b+c}\right)\)
(Theo BĐT cauchy với \(a,b,c>0\) )
\(\le\dfrac{1}{2}\left(\dfrac{\left(2a+2b+2c\right)^2}{4\left(a+b+c\right)}\right)=\dfrac{1}{2}.\left(\dfrac{6^2}{4.3}\right)=\dfrac{3}{2}\)
(theo BĐT cauchy schwarz)
Vậy Max P =\(\dfrac{3}{2}\Leftrightarrow a=b=c=1\)
15 tháng 8 2016
Ta có : \(\frac{bc}{\sqrt{3a+bc}}=\frac{bc}{\sqrt{\left(a+b+c\right)a+bc}}=\frac{bc}{\sqrt{a^2+ab+ac+bc}}=\frac{bc}{\sqrt{\left(a+b\right)\left(a+c\right)}}\)
Áp dụng bđt Cauchy , ta có : \(\frac{bc}{\sqrt{\left(a+b\right)\left(a+c\right)}}\le\frac{bc}{2}\left(\frac{1}{a+b}+\frac{1}{a+c}\right)\)
Tương tự : \(\frac{ac}{\sqrt{3b+ac}}=\frac{ac}{\sqrt{\left(a+b\right)\left(b+c\right)}}\le\frac{ac}{2}\left(\frac{1}{a+b}+\frac{1}{b+c}\right)\); \(\frac{ab}{\sqrt{3c+ab}}=\frac{ab}{\sqrt{\left(a+c\right)\left(b+c\right)}}\le\frac{ab}{2}\left(\frac{1}{a+c}+\frac{1}{b+c}\right)\)
\(\Rightarrow P=\frac{bc}{\sqrt{\left(a+b\right)\left(a+c\right)}}+\frac{ac}{\sqrt{\left(b+a\right)\left(b+c\right)}}+\frac{ab}{\sqrt{\left(a+c\right)\left(c+b\right)}}\)
\(\le\frac{1}{2}\left(\frac{ab}{a+c}+\frac{ab}{b+c}+\frac{bc}{a+b}+\frac{bc}{a+c}+\frac{ac}{a+b}+\frac{ac}{b+c}\right)\)
\(\Rightarrow P\le\frac{1}{2}\left(\frac{ab+bc}{a+c}+\frac{ab+ac}{b+c}+\frac{bc+ac}{a+b}\right)=\frac{1}{2}\left(a+b+c\right)=\frac{3}{2}\)
Suy ra : Max P \(=\frac{3}{2}\Leftrightarrow a=b=c=1\)
TN
15 tháng 8 2016
đây nhé Câu hỏi của Steffy Han - Toán lớp 8 | Học trực tuyến
8 tháng 1 2020
\(\left(a+b\right)\left(b+c\right)\left(c+a\right)+abc\)
\(=abc+a^2b+ab^2+a^2c+ac^2+b^2c+bc^2+abc+abc\)
\(=\left(a+b+c\right)\left(ab+bc+ca\right)\)( phân tích nhân tử các kiểu )
\(\Rightarrow\left(a+b\right)\left(b+c\right)\left(c+a\right)\ge\left(a+b+c\right)\left(ab+bc+ca\right)-abc\left(1\right)\)
\(a+b+c\ge3\sqrt[3]{abc};ab+bc+ca\ge3\sqrt[3]{a^2b^2c^2}\Rightarrow\left(a+b+c\right)\left(ab+bc+ca\right)\ge9abc\)
\(\Rightarrow-abc\ge\frac{-\left(a+b+c\right)\left(ab+bc+ca\right)}{9}\)
Khi đó:\(\left(a+b+c\right)\left(ab+bc+ca\right)-abc\)
\(\ge\left(a+b+c\right)\left(ab+bc+ca\right)-\frac{\left(a+b+c\right)\left(ab+bc+ca\right)}{9}\)
\(=\frac{8\left(a+b+c\right)\left(ab+bc+ca\right)}{9}\left(2\right)\)
Từ ( 1 ) và ( 2 ) có đpcm
