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\(c+ab=\left(a+b+c\right)c+ab=ac+cb+c^2+ab=\left(a+c\right)\left(b+c\right)\)
Tương tự : \(a+bc=\left(a+b\right)\left(a+c\right);c+ab=\left(c+a\right)\left(c+b\right)\)
\(P=\sqrt{\frac{ab}{\left(c+a\right)\left(c+b\right)}}+\sqrt{\frac{bc}{\left(a+b\right)\left(a+c\right)}}+\sqrt{\frac{ca}{\left(b+c\right)\left(b+a\right)}}\)
áp dụng bất đẳng tức cauchy :
\(\sqrt{\frac{ab}{\left(c+a\right)\left(c+b\right)}}\le\frac{1}{2}\left(\frac{a}{c+a}+\frac{b}{c+b}\right)\)
\(\sqrt{\frac{bc}{\left(a+b\right)\left(a+c\right)}}\le\frac{1}{2}\left(\frac{b}{a+b}+\frac{c}{a+c}\right)\)
\(\sqrt{\frac{ca}{\left(b+c\right)\left(b+a\right)}}\le\frac{1}{2}\left(\frac{c}{b+c}+\frac{a}{b+a}\right)\)
cộng vế theo vế
\(\Rightarrow P\le\frac{1}{2}\left(\frac{a}{a+c}+\frac{b}{c+b}+\frac{b}{a+b}+\frac{c}{a+c}+\frac{c}{b+c}+\frac{a}{b+a}\right)\)
\(\Leftrightarrow P\le\frac{1}{2}\left(\frac{a+c}{a+c}+\frac{b+c}{b+c}+\frac{a+b}{a+b}\right)=\frac{1}{2}\cdot3=\frac{3}{2}\)
dấu "=" xảy ra khi a=b=c=1/3
Có a+b+c=1 => c=(a+b+c).c=ac+bc+c2
\(\Rightarrow c+ab=ac+bc+c^2+ab=a\left(b+c\right)+c\left(b+c\right)=\left(b+c\right)\left(a+c\right)\)
\(\Rightarrow\sqrt{\frac{ab}{c+ab}}=\sqrt{\frac{ab}{\left(c+a\right)\left(c+b\right)}}\le\frac{\frac{a}{c+b}+\frac{b}{c+b}}{2}\)
Tương tự ta có \(\hept{\begin{cases}a+bc=\left(a+b\right)\left(a+c\right)\\b+ac=\left(b+a\right)\left(b+c\right)\end{cases}\Leftrightarrow\hept{\begin{cases}\sqrt{\frac{bc}{a+bc}}=\sqrt{\frac{bc}{\left(a+b\right)\left(a+c\right)}}\le\frac{\frac{b}{a+b}+\frac{c}{a+c}}{2}\\\sqrt{\frac{ca}{b+ca}}=\sqrt{\frac{ca}{\left(b+c\right)\left(b+a\right)}}\le\frac{\frac{c}{b+c}+\frac{a}{b+a}}{2}\end{cases}}}\)
\(\Rightarrow P\le\frac{\frac{b}{a+b}+\frac{c}{c+a}+\frac{c}{b+c}+\frac{a}{a+b}+\frac{a}{c+a}+\frac{b}{c+b}}{2}\)\(=\frac{\frac{a+c}{a+c}+\frac{c+b}{c+b}+\frac{a+b}{a+b}}{2}=\frac{3}{2}\)
Dấu "=" xảy ra khi \(a=b=c=\frac{1}{3}\)
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: \(a=b=c=1\)
Ta có \(\sqrt{\frac{ab}{c+ab}}=\sqrt{\frac{ab}{c\left(a+b+c\right)+ab}}=\sqrt{\frac{ab}{\left(c+b\right)\left(c+a\right)}}\le\frac{1}{2}\left(\frac{a}{c+a}+\frac{b}{a+b}\right)\)
Khi đó \(P\le\frac{1}{2}\left(\frac{a}{a+b}+\frac{b}{a+b}\right)+\frac{1}{2}\left(\frac{a}{a+c}+\frac{c}{a+c}\right)+\frac{1}{2}\left(\frac{b}{b+c}+\frac{c}{b+c}\right)=\frac{3}{2}\)
\(MaxP=\frac{3}{2}\)khi a=b=c=1/3
Cách 1:
Ta có: \(\sqrt{\frac{ab}{c+ab}}=\sqrt{\frac{ab}{c\left(a+b+c\right)+ab}}=\sqrt{\frac{ab}{\left(c+a\right)\left(c+b\right)}}\le\frac{1}{2}\left(\frac{a}{c+a}+\frac{b}{c+b}\right)\)
Tương tự với \(\sqrt{\frac{bc}{a+bc}},\sqrt{\frac{ca}{b+ca}}\)rồi cộng các vế lại với nhau ta sẽ có
\(P\le\frac{3}{2}\)
Dấu đẳng thức xảy ra khi \(a=b=c=\frac{1}{3}\)
Vậy....
https://hoc247.net/hoi-dap/toan-10/tim-gtln-cua-p-can-ab-c-ab-can-bc-a-bc-can-ca-b-ca--faq406508.html
BĐT Cô-si:
\(\sqrt{xy}\le\frac{1}{2}\left(x+y\right)\)
Bạn tách căn kia ra là nhìn thấy vấn đề:
\(\sqrt{\frac{ab}{\left(a+c\right)\left(b+c\right)}}=\sqrt{\left(\frac{a}{a+c}\right)\left(\frac{b}{b+c}\right)}\)
Bài 1: Ta có \(\left(\frac{a^2}{b}-a+b\right)+b^2=\frac{a^2-ab+b^2}{b}+b\ge2\sqrt{a^2-ab+b^2}\) (áp dụng Bất Đẳng Thức Cosi)
\(=\sqrt{a^2-ab+b^2}+\sqrt{\frac{3}{4}\left(a-b\right)^2+\frac{1}{4}\left(a+b\right)^2}\ge\sqrt{a^2-ab+b^2}+\frac{1}{2}\left(a+b\right)\)
\(\Rightarrow\frac{a^2}{b}-a+2b\ge\sqrt{a^2-ab+b^2}+\frac{1}{2}\left(a+b\right)\left(1\right)\)
Tương tự ta có \(\hept{\begin{cases}\frac{b^2}{c}-b+2c\ge\sqrt{b^2-bc+c^2}+\frac{1}{2}\left(b+c\right)\left(2\right)\\\frac{c^2}{a}-c+2a\ge\sqrt{c^2-ac+a^2}+\frac{1}{2}\left(a+c\right)\left(3\right)\end{cases}}\)
Từ (1) và (2) và (3) \(\Rightarrow\frac{a^2}{b}+\frac{b^2}{c}+\frac{c^2}{a}\ge\sqrt{a^2-ab+b^2}+\sqrt{b^2-bc+c^2}+\sqrt{c^2-ac+a^2}\)
Dấu "=" xảy ra khi a=b=c
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:
\(\frac{bc}{\sqrt{a+bc}}+\frac{ca}{\sqrt{b+ca}}+\frac{ab}{\sqrt{c+ab}}\le\frac{1}{2}\left(a+b+c\right)=\frac{1}{2}\)