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Áp dụng BĐT AM - GM
\(A=\left(a+1\right)\left(1+\frac{1}{b}\right)+\left(b+1\right)\left(1+\frac{1}{a}\right)\)
\(=\frac{a}{b}+\frac{b}{a}+a+\frac{1}{a}+b+\frac{1}{b}+2\)
\(=\frac{a}{b}+\frac{b}{a}+\left(a+\frac{1}{2a}\right)+\left(b+\frac{1}{2b}\right)+\frac{1}{2a}+\frac{1}{2b}+2\)
\(\ge2\sqrt{\frac{a}{b}.\frac{b}{a}}+2\sqrt{a.\frac{1}{2a}}+2\sqrt{b.\frac{1}{2b}}+2\sqrt{\frac{1}{2a}.\frac{1}{2b}}+2\)
\(=4+2\sqrt{2}+\frac{1}{\sqrt{ab}}\ge4+2\sqrt{2}+\frac{1}{\frac{\sqrt{2\left(a^2+b^2\right)}}{2}}\)
\(=4+3\sqrt{2}\)
Dấu " = " xảy ra khi \(a=b=\frac{1}{\sqrt{2}}\)
Ta co:\(1=a^2+b^2\ge\frac{\left(a+b\right)^2}{2}\Rightarrow a+b\le\sqrt{2}\)
Ta lai co:
\(A=\frac{a}{b}+\frac{b}{a}+\frac{1}{a}+\frac{1}{b}+a+b+2\)
\(=\left(\frac{a}{b}+\frac{b}{a}\right)+\left(\frac{1}{a}+2a\right)+\left(\frac{1}{b}+2b\right)-\left(a+b\right)+2\)
\(\ge2+2\sqrt{2}+2\sqrt{2}-\sqrt{2}+2=4+3\sqrt{2}\)
Dau '=' xay ra khi \(a=b=\frac{1}{\sqrt{2}}\)
Vay \(A_{min}=4+3\sqrt{2}\)khi \(a=b=\frac{1}{\sqrt{2}}\)
bài 1
ÁP dụng AM-GM ta có:
\(\frac{a^3}{b\left(2c+a\right)}+\frac{2c+a}{9}+\frac{b}{3}\ge3\sqrt[3]{\frac{a^3.\left(2c+a\right).b}{b\left(2c+a\right).27}}=a.\)
tương tự ta có:\(\frac{b^3}{c\left(2a+b\right)}+\frac{2a+b}{9}+\frac{c}{3}\ge b,\frac{c^3}{a\left(2b+c\right)}+\frac{2b+c}{9}+\frac{a}{3}\ge c\)
công tất cả lại ta có:
\(P+\frac{2a+b}{9}+\frac{2b+c}{9}+\frac{2c+a}{9}+\frac{a+b+c}{3}\ge a+b+c\)
\(P+\frac{2\left(a+b+c\right)}{3}\ge a+b+c\)
Thay \(a+b+c=3\)vào ta được":
\(P+2\ge3\Leftrightarrow P\ge1\)
Vậy Min là \(1\)
dấu \(=\)xảy ra khi \(a=b=c=1\)
Sửa: \(a;b>0\)
Áp dụng BĐT AM-GM ta có:
\(A=\left(a+1\right)\left(1+\dfrac{1}{b}\right)+\left(b+1\right)\left(1+\dfrac{1}{a}\right)\)
\(=\dfrac{a}{b}+\dfrac{b}{a}+a+\dfrac{1}{a}+b+\dfrac{1}{b}+2\)
\(=\dfrac{a}{b}+\dfrac{b}{a}+\left(a+\dfrac{1}{2a}\right)+\left(b+\dfrac{1}{2b}\right)+\dfrac{1}{2a}+\dfrac{1}{2b}+2\)
\(\ge2\sqrt{\dfrac{a}{b}\cdot\dfrac{b}{a}}+2\sqrt{a\cdot\dfrac{1}{2a}}+2\sqrt{b\cdot\dfrac{1}{2b}}+2\sqrt{\dfrac{1}{2a}\cdot\dfrac{1}{2b}}+2\)
\(=4+2\sqrt{2}+\dfrac{1}{\sqrt{ab}}\)\(\ge4+2\sqrt{2}+\dfrac{1}{\dfrac{\sqrt{2\left(a^2+b^2\right)}}{2}}\)
\(=4+3\sqrt{2}\)
Dấu \("="\) xảy ra khi \(a=b=\dfrac{1}{\sqrt{2}}\)
Ta co:
\(P\ge21\left(a^2+b^2+c^2\right)+12\left(a+b+c\right)^2+\frac{2017.9}{2}\)
\(=21\left(a^2+b^2+c^2\right)+12\left(a+b+c\right)^2+\frac{18153}{2}\)
\(\Leftrightarrow\frac{P}{\left(a+b+c\right)^2}\ge21\left[\left(\frac{a}{a+b+c}\right)^2+\left(\frac{b}{a+b+c}\right)^2+\left(\frac{c}{a+b+c}\right)^2\right]+12+\frac{\frac{18153}{2}}{\left(a+b+c\right)^2}\)
Dat \(\left(\frac{a}{a+b+c};\frac{b}{a+b+c};\frac{c}{a+b+c}\right)\rightarrow\left(x;y;z\right)\)
\(\Rightarrow x+y+z=1\)
\(\Rightarrow\left(a+b+c\right)^2=\frac{a^2}{x^2}\)
BDT tro thanh:
\(\frac{P}{\left(a+b+c\right)^2}\ge21\left(x^2+y^2+z^2\right)+12+\frac{18153}{2\left(a+b+c\right)^2}\)
\(\Leftrightarrow\frac{P}{\frac{a^2}{x^2}}\ge21\left(x^2+y^2+z^2\right)+12+\frac{18153}{2\left(a+b+c\right)^2}\ge21.\frac{\left(x+y+z\right)^2}{3}+12+\frac{18153}{8}\)
\(\Leftrightarrow\frac{x^2P}{a^2}\ge7+12+\frac{18153}{8}\)
Ta lai co:\(x=\frac{a}{a+b+c}\ge\frac{a}{2}\Rightarrow a^2\le4x^2\)
Suy ra:\(\frac{x^2P}{a^2}\ge\frac{x^2P}{4x^2}=\frac{P}{4}\)
\(\Rightarrow\frac{P}{4}\ge\frac{18503}{8}\)
\(\Leftrightarrow P\ge\frac{18503}{2}\)
Dau '=' xay ra khi \(a=b=c=\frac{2}{3}\)
Vay \(P_{min}=\frac{18503}{2}\)khi \(a=b=c=\frac{2}{3}\)
\(2=\left|a-1\right|+\left|b-1\right|\ge\left|a-1+b-1\right|=\left|a+b-2\right|\)
Với \(a+b-2\ge0\Leftrightarrow a+b\ge2\) thì:
\(a+b-2\le2\Leftrightarrow a+b\le4\Rightarrow\left|a+b-1\right|\le\left|4-1\right|=3\)
Dấu "=" xảy ra \(\left\{{}\begin{matrix}2\le a+b\le4\\a+b=4\\\left(a-1\right)\left(b-1\right)\ge0\end{matrix}\right.\Leftrightarrow\left\{{}\begin{matrix}a\ge1,b\ge1\\a+b=4\end{matrix}\right.\)
Với \(a+b-2\le0\Leftrightarrow a+b\le2\) thì:
\(-a-b+2\le2\Leftrightarrow a+b\ge0\Rightarrow\left|a+b-1\right|\ge\left|0-1\right|=1\)
Dấu "=" xảy ra \(\Leftrightarrow\left\{{}\begin{matrix}a+b\le2\\a+b=0\\\left(a-1\right)\left(b-1\right)\ge0\end{matrix}\right.\Leftrightarrow\left\{{}\begin{matrix}a\le1,b\le1\\a+b=0\end{matrix}\right.\)
Vậy GTNN của \(\left|a+b-1\right|\) là 1 khi \(\left\{{}\begin{matrix}a+b=0\\a\le1,b\le1\end{matrix}\right.\)
GTLN của $\left|a+b-1\right|$ là 3 khi $\hept{\begin{matrix}a\ge 1,b\ge 1\\a+b=4\end{matrix}}$
\(\left(\sqrt{a}+1\right)\left(\sqrt{b}+1\right)=4\Leftrightarrow\sqrt{ab}+\sqrt{a}+\sqrt{b}=3\)
\(\text{Ta có:}M\ge a+b\Rightarrow2M+2\ge a+b+a+1+b+1\ge2\left(\sqrt{ab}+\sqrt{a}+\sqrt{b}\right)\left(\text{theo cô si}\right)=6\)
\(\Rightarrow M\ge2\left(\text{dấu "=" xảy ra khi:}a=b=1\right)\)