Cho \(a,b>0\) và \(a^2b+ab^2+ab=a^2+b^2\). Tìm max P \(=\) \(\frac{1}{a}\sqrt{1+\frac{a}{b}}+\frac{1}{b}\sqrt{1+\frac{b}{a}}\)
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VT
12 tháng 7 2018
Ta có \(\sqrt{a^2-ab+b^2}=\sqrt{\frac{1}{4}\left(a+b\right)^2+\frac{3}{4}\left(a-b\right)^2}\ge\sqrt{\frac{1}{4}\left(a+b\right)^2}=\frac{1}{2}\left(a+b\right)\)
=> \(\frac{1}{\sqrt{a^2-ab+b^2}}\le\frac{1}{\frac{1}{2}\left(a+b\right)}=\frac{2}{a+b}\le\frac{1}{2}\left(\frac{1}{a}+\frac{1}{b}\right)\)
Chứng minh tương tự, rồi cộng lại, ta có
A\(\le\frac{1}{a}+\frac{1}{b}+\frac{1}{c}=3\)
dấu = xảy ra <=> a=b=c=1
^_^
24 tháng 9 2020
1) \(VT=\frac{\sqrt{a}+\sqrt{b}}{2\left(\sqrt{a}-\sqrt{b}\right)}-\frac{\sqrt{a}-\sqrt{b}}{2\left(\sqrt{a}+\sqrt{b}\right)}+\frac{2b}{\left(\sqrt{a}+\sqrt{b}\right)\left(\sqrt{a}-\sqrt{b}\right)}\)
\(=\frac{\left(\sqrt{a}+\sqrt{b}\right)^2-\left(\sqrt{a}-\sqrt{b}\right)^2+4b}{2\left(\sqrt{a}+\sqrt{b}\right)\left(\sqrt{a}-\sqrt{b}\right)}\)\(=\frac{a+2\sqrt{ab}+b-a+2\sqrt{ab}-b+4b}{2\left(\sqrt{a}+\sqrt{b}\right)\left(\sqrt{a}-\sqrt{b}\right)}\)
\(=\frac{4\sqrt{ab}+4b}{2\left(\sqrt{a}+\sqrt{b}\right)\left(\sqrt{a}-\sqrt{b}\right)}\)
\(=\frac{4\sqrt{b}\left(\sqrt{a}+\sqrt{b}\right)}{2\left(\sqrt{a}+\sqrt{b}\right)\left(\sqrt{a}-\sqrt{b}\right)}=\frac{2\sqrt{b}}{\sqrt{a}-\sqrt{b}}=VP\)(ĐPCM)
2) \(VT=\text{[}\frac{\left(\sqrt{a}+\sqrt{b}\right)\left(a+b-\sqrt{ab}\right)}{\left(\sqrt{a}+\sqrt{b}\right)}-\sqrt{ab}\text{]}.\frac{\left(\sqrt{a}+\sqrt{b}\right)^2}{\left(a-b\right)^2}\)
\(=\frac{\left(a+b-\sqrt{ab}-\sqrt{ab}\right)\left(\sqrt{a}+\sqrt{b}\right)^2}{\left(a-b\right)^2}\)\(=\frac{\left(\sqrt{a}-\sqrt{b}\right)^2\left(\sqrt{a}+\sqrt{b}\right)^2}{\left(a-b\right)^2}=\frac{\left(a-b\right)^2}{\left(a-b\right)^2}=1=VP\)(ĐPCM)
4) \(VT=\left(1+\frac{a+\sqrt{a}}{\sqrt{a}+1}\right)\left(1-\frac{a-\sqrt{a}}{\sqrt{a}-1}\right)\)\(=\left(1+\frac{\sqrt{a}\left(\sqrt{a}+1\right)}{\sqrt{a}+1}\right)\left(1-\frac{\sqrt{a}\left(\sqrt{a}-1\right)}{\sqrt{a}-1}\right)\)
\(=\left(1+\sqrt{a}\right)\left(1-\sqrt{a}\right)=1-a=VP\)(ĐPCM)
31 tháng 1 2019
Ta có : \(\frac{a}{\sqrt{bc\left(1+a^2\right)}}=\frac{a}{\sqrt{bc+a.abc}}=\frac{a}{\sqrt{bc+a\left(a+b+c\right)}}\)
\(=\frac{a}{\sqrt{bc+a^2+ab+ac}}\)
\(=\frac{a}{\sqrt{\left(a+b\right)\left(a+c\right)}}\)
Áp dụng bđt Cô-si ngược ta có
\(\frac{a}{\sqrt{bc\left(1+a^2\right)}}=\frac{a}{\sqrt{\left(a+b\right)\left(a+c\right)}}\le\frac{1}{2}\left(\frac{a}{a+b}+\frac{a}{a+c}\right)\)
C/m tương tự được \(\frac{b}{\sqrt{ca\left(1+b^2\right)}}\le\frac{1}{2}\left(\frac{b}{a+b}+\frac{b}{b+c}\right)\)
\(\frac{c}{\sqrt{ab\left(1+c^2\right)}}\le\frac{1}{2}\left(\frac{c}{a+c}+\frac{c}{b+c}\right)\)
Cộng 3 vế của các bđt trên lại ta được
\(A\le\frac{1}{2}\left(\frac{a}{a+b}+\frac{b}{a+b}+\frac{a}{a+c}+\frac{c}{a+c}+\frac{b}{b+c}+\frac{c}{b+c}\right)\)
\(=\frac{3}{2}\)
Dấu "=" xảy ra \(\Leftrightarrow\hept{\begin{cases}a+b+c=abc\\a=b=c\end{cases}}\Leftrightarrow\hept{\begin{cases}3a=a^3\\a=b=c\end{cases}}\)
\(\Leftrightarrow\hept{\begin{cases}a^3-3a=0\\a=b=c\end{cases}}\)
\(\Leftrightarrow\hept{\begin{cases}a\left(a^2-3\right)=0\\a=b=c\end{cases}}\)
\(\Leftrightarrow a=b=c=\sqrt{3}\left(a,b,c>0\right)\)
Vậy \(A_{max}=\frac{3}{2}\Leftrightarrow x=y=z=\sqrt{3}\)
NV
Nguyễn Việt Lâm
Giáo viên
24 tháng 9 2019
Chỉ tìm được khi a;b;c dương, còn ko có điều kiện dương thì chịu thua :(
NV
Nguyễn Việt Lâm
Giáo viên
3 tháng 6 2020
\(\frac{1}{a}+\frac{1}{b}+\frac{1}{a}+\frac{1}{c}+\frac{1}{b}+\frac{1}{c}\ge4\left(\frac{1}{a+b}+\frac{1}{a+c}+\frac{1}{b+c}\right)\ge2\)
\(\Rightarrow\frac{1}{a}+\frac{1}{b}+\frac{1}{c}\ge1\)
Đặt \(\left(\frac{1}{a};\frac{1}{b};\frac{1}{c}\right)=\left(x;y;z\right)\Rightarrow x+y+z\ge1\)
\(P=\sqrt{x^2+2y^2}+\sqrt{y^2+2z^2}+\sqrt{z^2+2x^2}\)
\(\Rightarrow P\ge\sqrt{\frac{\left(x+2y\right)^2}{3}}+\sqrt{\frac{\left(y+2z\right)^2}{3}}+\sqrt{\frac{\left(z+2x\right)^2}{3}}\)
\(\Rightarrow P\ge\frac{1}{\sqrt{3}}\left(3x+3y+3z\right)\ge\frac{3}{\sqrt{3}}=\sqrt{3}\)
Dấu "=" xảy ra khi \(x=y=z=\frac{1}{3}\) hay \(a=b=c=3\)
\(\frac{1}{2}\left(a+b\right)^2\le a^2+b^2=ab\left(a+b\right)+ab\le ab\left(a+b\right)+\frac{1}{4}\left(a+b\right)^2\)
\(\Rightarrow\frac{1}{2}\left(a+b\right)^2\le ab\left(a+b\right)+\frac{1}{4}\left(a+b\right)^2\)
\(\Rightarrow\frac{1}{4}\left(a+b\right)^2\le ab\left(a+b\right)\Rightarrow a+b\le4ab\)
\(\Rightarrow\frac{a+b}{ab}\le4\)
\(P=\frac{\sqrt{b\left(a+b\right)}}{ab}+\frac{\sqrt{a\left(a+b\right)}}{ab}=\frac{1}{2\sqrt{2}}\left(\frac{2\sqrt{2b\left(a+b\right)}+2\sqrt{2a\left(a+b\right)}}{ab}\right)\)
\(P\le\frac{1}{2\sqrt{2}}\left(\frac{2b+a+b+2a+a+b}{ab}\right)=\sqrt{2}\left(\frac{a+b}{ab}\right)\le4\sqrt{2}\)
\(P_{max}=4\sqrt{2}\) khi \(a=b=\frac{1}{2}\)
@Nguyễn Lê Phước Thịnh