tìm giá trị nhỏ nhất của biểu thucM ,biết
M=\(\left(1+a\right)\left(1+\frac{1}{b}\right)+\left(1+b\right)\left(1+\frac{1}{a}\right)\)) với a>0,b>0 va \(^{a^{2+}b^{2=1}}\)
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![](https://rs.olm.vn/images/avt/0.png?1311)
1. a) Ta có: M = |x + 15/19| \(\ge\)0 \(\forall\)x
Dấu "=" xảy ra <=> x + 15/19 = 0 <=> x = -15/19
Vậy MinM = 0 <=> x = -15/19
b) Ta có: N = |x - 4/7| - 1/2 \(\ge\)-1/2 \(\forall\)x
Dấu "=" xảy ra <=> x - 4/7 = 0 <=> x = 4/7
Vậy MinN = -1/2 <=> x = 4/7
2a) Ta có: P = -|5/3 - x| \(\le\)0 \(\forall\)x
Dấu "=" xảy ra <=> 5/3 - x = 0 <=> x = 5/3
Vậy MaxP = 0 <=> x = 5/3
b) Ta có: Q = 9 - |x - 1/10| \(\le\)9 \(\forall\)x
Dấu "=" xảy ra <=> x - 1/10 = 0 <=> x = 1/10
Vậy MaxQ = 9 <=> x = 1/10
![](https://rs.olm.vn/images/avt/0.png?1311)
\(A=\left(a+b+1\right)\left(a^2+b^2\right)+\frac{4}{a+b}+1-1\ge\left(a+b+1\right)2\sqrt{\left(ab\right)^2}+\frac{\left(2+1\right)^2}{a+b+1}-1\)
\(=2\left(a+b+1\right)+\frac{9}{a+b+1}-1\ge2\sqrt{ab}+1+2\sqrt{\frac{9\left(a+b+1\right)}{a+b+1}}-1\ge2+6=8\)
Dấu "=" xảy ra \(\Leftrightarrow\)\(\hept{\begin{cases}a^2=b^2\left(1\right)\\\frac{2}{a+b}=1\left(2\right)\\a+b+1=\frac{9}{a+b+1}\left(3\right)\end{cases}}\)
pt \(\left(1\right)\)\(\Leftrightarrow\)\(a=b\) ( vì a, b > 0 )
pt \(\left(2\right)\)\(\Leftrightarrow\)\(a=b=1\)
pt \(\left(3\right)\)\(\Leftrightarrow\)\(\left(a+b+1\right)^2=9\)\(\Leftrightarrow\)\(a+b+1=3\) ( đúng vì \(a=b=1\) )
Vậy GTNN của \(A\) là \(8\) khi \(a=b=1\)
Chúc bạn học tốt ~
![](https://rs.olm.vn/images/avt/0.png?1311)
Ta có : \(ab+bc+ca=2abc\)
\(\Leftrightarrow\frac{1}{a}+\frac{1}{b}+\frac{1}{c}=2\)
Đặt \(\hept{\begin{cases}x=\frac{1}{a}\\y=\frac{1}{b}\\z=\frac{1}{c}\end{cases}}\)
\(\Rightarrow\hept{\begin{cases}x+y+z=2\\P=\frac{x^3}{\left(2-x\right)^2}+\frac{y^3}{\left(2-y\right)^3}+\frac{z^3}{\left(2-z\right)^2}\end{cases}}\)
Áp dụng bất đẳng thức Cauchy - Schwarz
\(\Rightarrow\frac{x^3}{\left(2-x\right)^2}+\frac{2-x}{8}+\frac{2-x}{8}\ge3\sqrt[3]{\frac{x^3}{64}}=\frac{3x}{4}\)
Tương tự ta có :
\(\hept{\begin{cases}\frac{y^3}{\left(2-y\right)^2}+\frac{2-y}{8}+\frac{2-y}{8}\ge\frac{3y}{4}\\\frac{z^3}{\left(2-z\right)^2}+\frac{2-z}{8}+\frac{2-z}{8}\ge\frac{3z}{8}\end{cases}}\)
\(\Rightarrow P+\frac{12-2\left(x+y+z\right)}{8}\ge\frac{3}{4}\left(x+y+z\right)\)
\(\Rightarrow P\ge\frac{1}{12}\)
Dấu " = " xảy ra khi \(x=y=z=\frac{2}{3}\)
![](https://rs.olm.vn/images/avt/0.png?1311)
Ta có : \(ab+bc+ca=2abc\)
\(\Leftrightarrow\frac{1}{a}+\frac{1}{b}+\frac{1}{c}=2\)
Đặt \(\hept{\begin{cases}x=\frac{1}{a}\\y=\frac{1}{b}\\z=\frac{1}{c}\end{cases}}\)
\(\Rightarrow\hept{\begin{cases}x+y+z=2\\P=\frac{x^3}{\left(2-x\right)^2}+\frac{y^3}{\left(2-y\right)^3}+\frac{z^3}{\left(2-z^2\right)}\end{cases}}\)
Áp dụng bất đẳng thức Cauchy - Schwarz
\(\Rightarrow\frac{x^3}{\left(2-x\right)^2}+\frac{2-x}{8}+\frac{2-x}{8}\ge3\sqrt[3]{\frac{x^3}{64}}=\frac{3x}{4}\)
Tương tự ta có : \(\hept{\begin{cases}\frac{y^3}{\left(2-y\right)^2}+\frac{2-y}{8}+\frac{2-y}{8}\ge\frac{3y}{4}\\\frac{z^3}{\left(2-z\right)^2}+\frac{2-z}{8}+\frac{2-z}{8}\ge\frac{3z}{8}\end{cases}}\)
\(\Rightarrow P+\frac{12-2\left(x+y+z\right)}{8}\ge\frac{3}{4}\left(x+y+z\right)\)
\(\Rightarrow P\ge\frac{1}{2}\)
Dấu "=" xảy ra khi \(x=y=z=\frac{2}{3}\)
Ta có
\(M=\left(1+a\right)\left(1+\frac{1}{b}\right)+\left(1+b\right)\left(1+\frac{1}{a}\right)=2+\frac{a}{b}+\frac{b}{a}+a+b+\frac{1}{a}+\frac{1}{b}\)
\(\ge2+2+a+b+\frac{4}{a+b}\)
\(=4+a+b+\frac{2}{a+b}+\frac{2}{a+b}\)
\(\ge4+2\sqrt{\left(a+b\right).\frac{2}{\left(a+b\right)}}+\frac{2}{\sqrt{2\left(a^2+b^2\right)}}\)
\(=4+2\sqrt{2}+\sqrt{2}=4+3\sqrt{2}\)