Cho a,b,c>0 và a+b+c=1. Tìm giá trị lớn nhất của P = \(\frac{a}{1+a}+\frac{b}{1+b}+\frac{c}{1+c}\)
Hãy nhập câu hỏi của bạn vào đây, nếu là tài khoản VIP, bạn sẽ được ưu tiên trả lời.
![](https://rs.olm.vn/images/avt/0.png?1311)
![](https://rs.olm.vn/images/avt/0.png?1311)
Ta có: \(\frac{1}{a}+\frac{1}{b}+\frac{4}{c}\ge\frac{4}{a+b}+\frac{4}{c}=4\left(\frac{1}{a+b}+\frac{1}{c}\right)\ge4\frac{4}{a+b+c}=4.\frac{4}{6}=\frac{8}{3}\)
\(\Rightarrow-\left(\frac{1}{a}+\frac{1}{b}+\frac{4}{c}\right)\le\frac{-8}{3}\)
\(\Rightarrow M=1-\frac{1}{a}+1-\frac{1}{b}+1-\frac{4}{c}\)
\(=3-\left(\frac{1}{a}+\frac{1}{b}+\frac{4}{c}\right)\le3-\frac{8}{3}=\frac{1}{3}\)
\(\Rightarrow M\le\frac{1}{3}\)
Dấu '=' xảy ra \(\Leftrightarrow\hept{\begin{cases}a=b\\a+b=c\\a+b+c=6\end{cases}\Leftrightarrow\hept{\begin{cases}a=b=\frac{3}{2}\\c=3\end{cases}}}\)
Vậy GTLN của M là 1/3
![](https://rs.olm.vn/images/avt/0.png?1311)
Với 2 số x,y > 0 Theo Cauchy ta có: \(\frac{x+y}{2}\ge\sqrt{xy}\Rightarrow\frac{\left(x+y\right)^2}{4}\ge xy\Rightarrow\frac{x+y}{xy}\ge\frac{4}{x+y}\)
\(\Rightarrow\frac{1}{x}+\frac{1}{y}\ge\frac{4}{x+y}^{\left(1\right)}\)
\(P=\frac{a-1}{a}+\frac{b-1}{b}+\frac{c-4}{c}=1-\frac{1}{a}+1-\frac{1}{b}+1-\frac{4}{c}\)
\(=3-\left(\frac{1}{a}+\frac{1}{b}+\frac{4}{c}\right)\)
Áp dụng (1) ta có:\(\frac{1}{a}+\frac{1}{b}+\frac{4}{c}\ge\frac{4}{a+b}+\frac{4}{c}=4\left(\frac{1}{a+b}+\frac{1}{c}\right)\ge4\cdot\frac{4}{a+b+c}=\frac{16}{6}=\frac{8}{3}\)
\(\Rightarrow3-\left(\frac{1}{a}+\frac{1}{b}+\frac{4}{c}\right)\le3-\frac{8}{3}=\frac{1}{3}\)
Đẳng thức xảy ra khi a=b và (a+b)=c hay a=b=1,5 và c=3.
![](https://rs.olm.vn/images/avt/0.png?1311)
đặt \(a+b=x,b+c=y;c+a=z\)
ta có \(\frac{1}{x+1}+\frac{1}{y+1}+\frac{1}{z+1}=1\Rightarrow3-\frac{1}{x+1}-\frac{1}{y+1}-\frac{1}{z+1}=1\) \(\)
=> \(\frac{x}{x+1}+\frac{y}{y+1}+\frac{z}{z+1}=1\)
=> \(\frac{y}{y+1}+\frac{z}{z+1}=1-\frac{x}{x+1}=\frac{1}{x+1}\)
Áp dụng bđt cô si ta có \(\frac{y}{y+1}+\frac{z}{z+1}\ge2\sqrt{\frac{yz}{\left(y+1\right)\left(z+1\right)}}\)
=> \(\frac{1}{x+1}\ge2\sqrt{\frac{yz}{\left(y+1\right)\left(z+1\right)}}\)
tương tự ta có
\(\frac{1}{y+1}\ge2\sqrt{\frac{zx}{\left(z+1\right)\left(x+1\right)}}\)
\(\frac{1}{z+1}\ge2\sqrt{\frac{xy}{\left(x+1\right)\left(y+1\right)}}\)
nhân từng vế của 3 bđt cùng chièu ta có
\(\frac{1}{x+1}.\frac{1}{y+1}.\frac{1}{z+1}\ge8\sqrt{\frac{x^2y^2z^2}{\left(x+1\right)^2\left(y+1\right)^2\left(z+1\right)^2}}=8.\frac{xyz}{\left(x+1\right)\left(y+1\right)\left(z+1\right)}\)
=> \(1\ge8xyz\Rightarrow xyz\le\frac{1}{8}\Rightarrow\left(a+b\right)\left(b+c\right)\left(c+a\right)\le\frac{1}{8}\)
![](https://rs.olm.vn/images/avt/0.png?1311)
\(P=\frac{a-1}{a}+\frac{b-1}{b}+\frac{c-4}{c}=\frac{a}{a}-\frac{1}{a}+\frac{b}{b}-\frac{1}{b}+\frac{c}{c}-\frac{4}{c}\)
=> \(P=3-\left(\frac{1}{a}+\frac{1}{b}+\frac{4}{c}\right)\)(1)
Ta lại có: \(\left(\sqrt{a}-\sqrt{b}\right)^2\ge0< =>a+b-2\sqrt{ab}\ge0=>\frac{\left(a+b\right)^2}{4}\ge ab\)
<=> \(\frac{a+b}{ab}\ge\frac{4}{a+b}< =>\frac{1}{a}+\frac{1}{b}\ge\frac{4}{a+b}\)
=> \(\frac{1}{a}+\frac{1}{b}+\frac{4}{c}\ge\frac{4}{a+b}+\frac{4}{c}=4\left(\frac{1}{a+b}+\frac{1}{c}\right)\ge4\left(\frac{4}{a+b+c}\right)\)
=> \(\frac{1}{a}+\frac{1}{b}+\frac{4}{c}\ge4\left(\frac{4}{6}\right)=\frac{16}{6}=\frac{8}{3}\)(Do a+b+c=6 theo gt)
Thay vào (1), suy ra:
\(P=3-\left(\frac{1}{a}+\frac{1}{b}+\frac{4}{c}\right)\le3-\frac{8}{3}=\frac{1}{3}\)
=> GTLL của P là: \(P=\frac{1}{3}\)
Dấu '=' xảy ra khi a=b và a+b=c => c=3; a=b=1,5
![](https://rs.olm.vn/images/avt/0.png?1311)
Cần chứng minh: \(\sqrt{a^2-ab+b^2}\ge\frac{1}{2}\left(a+b\right)\)
Thật vậy: \(\sqrt{a^2-ab+b^2}\ge\frac{1}{2}\left(a+b\right)^2\Leftrightarrow4\left(a^2-ab+b^2\right)\ge\left(a+b\right)^2\)
\(\Leftrightarrow4a^2-4ab+4b^2-a^2-b^2-2ab\ge0\Leftrightarrow3\left(a^2+b^2-2ab\right)\ge0\Leftrightarrow3\left(a-b\right)^2\ge0\)(đúng)
Áp dụng:\(P=\frac{1}{\sqrt{a^2-ab+b^2}}+\frac{1}{\sqrt{b^2-bc+c^2}}+\frac{1}{\sqrt{c^2-ac+a^2}}\)
\(\le\frac{1}{\frac{1}{2}\left(a+b\right)}+\frac{1}{\frac{1}{2}\left(b+c\right)}+\frac{1}{\frac{1}{2}\left(c+a\right)}=2\left(\frac{1}{a+b}+\frac{1}{b+c}+\frac{1}{a+c}\right)=3\)
Dấu "=" xảy ra khi: \(a=b=c=1\)
![](https://rs.olm.vn/images/avt/0.png?1311)
\(b^4+c^4\ge bc\left(b^2+c^2\right)\)vì \(\left(b-c\right)^2\left(b^2+bc+c^2\right)\ge0\)
\(\Rightarrow T\le\frac{a}{\frac{b^2+c^2}{a}+a}+\frac{b}{\frac{a^2+c^2}{b}+b}+\frac{c}{\frac{a^2+b^2}{c}+c}=1\)
![](https://rs.olm.vn/images/avt/0.png?1311)
Ta có:
\(P=\frac{ab}{\sqrt{c+ab}}+\frac{bc}{\sqrt{a+bc}}+\frac{ca}{\sqrt{b+ca}}\)
\(=\frac{ab}{\sqrt{1-a-b+ab}}+\frac{bc}{\sqrt{1-b-c+bc}}+\frac{ca}{\sqrt{1-a-c+ca}}\)
\(=\frac{ab}{\sqrt{\left(1-a\right)\left(1-b\right)}}+\frac{bc}{\sqrt{\left(1-b\right)\left(1-c\right)}}+\frac{ca}{\sqrt{\left(1-c\right)\left(1-a\right)}}\)
\(\le\frac{a^2}{2\left(1-a\right)}+\frac{b^2}{2\left(1-b\right)}+\frac{b^2}{2\left(1-b\right)}+\frac{c^2}{2\left(1-c\right)}+\frac{c^2}{2\left(1-c\right)}+\frac{a^2}{2\left(1-a\right)}\)
\(=-\left(\frac{a^2}{a-1}+\frac{b^2}{b-1}+\frac{c^2}{c-1}\right)\)
\(\le-\frac{\left(a+b+c\right)^2}{a+b+c-3}=\frac{1}{3-1}=\frac{1}{2}\)
Vậy GTLN là \(P=\frac{1}{2}\) khi \(a=b=c=\frac{1}{3}\)
Biến đổi một chút, ta có:\(\frac{bc}{\sqrt{a+bc}}=\frac{bc}{\sqrt{a\left(a+b+c\right)+bc}}\)
\(=\sqrt{\frac{bc}{a+bc}}\cdot\sqrt{\frac{bc}{c+a}}\le\frac{1}{2}\left(\frac{bc}{a+b}+\frac{bc}{a+c}\right)\)
Tương tự cho 2 BĐT còn lại ta có:
\(\frac{ca}{\sqrt{b+ca}}\le\frac{1}{2}\left(\frac{ca}{a+b}+\frac{ca}{b+c}\right);\frac{ab}{\sqrt{c+ab}}\le\frac{1}{2}\left(\frac{ab}{a+c}+\frac{ab}{a+b}\right)\)
Cộng ba bất đẳng thức trên lại theo vế, ta có:
\(\frac{bc}{\sqrt{a+bc}}+\frac{ca}{\sqrt{b+ca}}+\frac{ab}{\sqrt{c+ab}}\le\frac{1}{2}\left(a+b+c\right)=\frac{1}{2}\)
\(\Sigma\frac{a}{1+a}=\Sigma\frac{1}{16}a\left(\frac{16}{a+\frac{1}{3}+\frac{1}{3}+\frac{1}{3}}\right)\le\Sigma\frac{1}{16}a\left(\frac{1}{a}+9\right)=\frac{3}{4}\)