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\(P=\frac{ab}{\sqrt{\left(c+a\right)\left(b+c\right)}}+\frac{bc}{\sqrt{\left(c+a\right)\left(a+b\right)}}+\frac{ca}{\sqrt{\left(b+c\right)\left(a+b\right)}}\)
thử dùng cô si đi
\(VT=\sum\frac{ab}{\sqrt{\left(a+b+c\right)c+ab}}=\sum\frac{ab}{\sqrt{\left(b+c\right)\left(c+a\right)}}\le\sum\frac{ab}{2}\left(\frac{1}{b+c}+\frac{1}{c+a}\right)\)
\(=\frac{1}{2}\left[\frac{ab+ca}{b+c}+\frac{ab+bc}{c+a}+\frac{bc+ca}{a+b}\right]=\frac{1}{2}\left(a+b+c\right)=1\)
Ta có \(\frac{\sqrt{ab^2c^3}}{b+c}\le\frac{\sqrt{ab^2c^3}}{2\sqrt{bc}}=\frac{1}{2}.\sqrt{ac.bc}\)
Mà \(\frac{1}{2}\sqrt{ac.cb}\le\frac{1}{4}\left(ac+cb\right)\)\(\Rightarrow\frac{\sqrt{ab^2c^3}}{b+c}\le\frac{1}{4}\left(ac+bc\right)\)
Tương tự cộng lại, ta có
\(\frac{\sqrt{ab^2c^3}}{b+c}+\frac{\sqrt{bc^2a^3}}{c+a}+\frac{\sqrt{ca^2b^3}}{a+b}\le\frac{1}{2}\left(ab+bc+ca\right)\)
Mà \(ab+bc+ca\le\frac{1}{3}\left(a+b+c\right)^2=3\Rightarrow\frac{\sqrt{ab^2c^3}}{b+c}+...\le\frac{3}{2}\)
dấu = xảy ra <=> a=b=c=1
^.^
\(\dfrac{ab}{\sqrt{ab+2c}}=\dfrac{ab}{\sqrt{ab+\left(a+b+c\right)c}}=\dfrac{ab}{\sqrt{\left(a+c\right)\left(b+c\right)}}=ab\cdot\sqrt{\dfrac{1}{a+b}\cdot\dfrac{1}{b+c}}\le ab\cdot\dfrac{1}{2}\left(\dfrac{1}{a+b}+\dfrac{1}{b+c}\right)=\dfrac{1}{2}\left(\dfrac{ab}{a+b}+\dfrac{ab}{b+c}\right)\)
CMTT: \(\dfrac{bc}{\sqrt{bc+2a}}\le\dfrac{1}{2}\left(\dfrac{bc}{a+b}+\dfrac{bc}{a+c}\right);\dfrac{ac}{\sqrt{ac+2b}}\le\dfrac{1}{2}\left(\dfrac{ac}{b+c}+\dfrac{ac}{b+a}\right)\)
\(\Leftrightarrow P\le\dfrac{1}{2}\left(\dfrac{ab}{c+a}+\dfrac{ab}{c+b}+\dfrac{bc}{b+a}+\dfrac{bc}{c+a}+\dfrac{ac}{b+c}+\dfrac{ac}{b+c}\right)\\ \Leftrightarrow P\le\dfrac{1}{2}\left[\dfrac{b\left(a+c\right)}{a+c}+\dfrac{a\left(b+c\right)}{b+c}+\dfrac{c\left(a+b\right)}{a+b}\right]=\dfrac{1}{2}\left(a+b+c\right)=1\)
Dấu \("="\Leftrightarrow a=b=c=\dfrac{2}{3}\)
\(VT=\sqrt{\frac{ab+2c^2}{a^2+ab+b^2}}+\sqrt{\frac{bc+2a^2}{b^2+bc+c^2}}+\sqrt{\frac{ca+2b^2}{c^2+ca+a^2}}\)
\(=\frac{ab+2c^2}{\sqrt{\left(a^2+ab+b^2\right)\left(ab+2c^2\right)}}+\frac{bc+2a^2}{\sqrt{\left(b^2+bc+c^2\right)\left(bc+2a^2\right)}}+\frac{ca+2b^2}{\sqrt{\left(c^2+ca+a^2\right)\left(ca+2b^2\right)}}\)
\(\ge\frac{2\left(ab+2c^2\right)}{a^2+b^2+2c^2+2ab}+\frac{2\left(bc+2a^2\right)}{2a^2+b^2+c^2+2bc}+\frac{2\left(ca+2b^2\right)}{a^2+2b^2+c^2+2ca}\)
\(\ge\frac{ab+2c^2}{a^2+b^2+c^2}+\frac{bc+2a^2}{a^2+b^2+c^2}+\frac{ca+2b^2}{a^2+b^2+c^2}=ab+bc+ca+2\left(a^2+b^2+c^2\right)\)
\(=2+ab+bc+ca=VP\) (Do a2 + b2 + c2 = 1) => ĐPCM.
Dấu "=" xảy ra <=> \(a=b=c=\frac{1}{\sqrt{3}}.\)
chăc là .............................. điền đi sẽ biếc a you ok ?
Với \(a^2+b^2+c^2=1\), ta có: \(\Sigma\sqrt{\frac{ab+2c^2}{1+ab-c^2}}=\Sigma\sqrt{\frac{ab+2c^2}{a^2+b^2+c^2+ab-c^2}}\)
\(=\Sigma\sqrt{\frac{ab+2c^2}{a^2+b^2+ab}}=\Sigma\frac{ab+2c^2}{\sqrt{\left(ab+2c^2\right)\left(a^2+b^2+ab\right)}}\)
\(\ge\Sigma\frac{ab+2c^2}{\frac{\left(ab+2c^2\right)+\left(a^2+b^2+ab\right)}{2}}=\Sigma\frac{ab+2c^2}{\frac{\left(a^2+b^2\right)+2ab+2c^2}{2}}\)
\(\ge\text{}\Sigma\text{}\frac{ab+2c^2}{\frac{\left(a^2+b^2\right)+\left(a^2+b^2\right)+2c^2}{2}}=\Sigma\frac{ab+2c^2}{\frac{2\left(a^2+b^2+c^2\right)}{2}}\)
\(=\Sigma\left(ab+2c^2\right)=2\left(a^2+b^2+c^2\right)+ab+bc+ca\)
\(=2+ab+bc+ca\)
Đẳng thức xảy ra khi \(a=b=c=\frac{1}{\sqrt{3}}\)
ta có \(\sqrt{\frac{ab+2c^2}{1+ab-c^2}}=\frac{ab+2c^2}{\sqrt{1+ab-c^2}.\sqrt{ab+2c^2}}=\frac{ab+2c^2}{\sqrt{1+ab-c^2}\sqrt{ab+2c^2}}\)
Áp dụng bất đẳng thức cô si ta có
\(\sqrt{ab+1-c^2}\sqrt{ab+2c^2}\le\frac{1}{2}\left(ab+1-c^2+ab+2c^2\right)=\frac{1}{2}\left(2ab+1+c^2\right)\)
=\(\frac{1}{2}\left(2ab+a^2+b^2+2c^2\right)=\frac{1}{2}\left[\left(a+b\right)^2+2c^2\right]\le\frac{1}{2}\left(2a^2+2b^2+2c^2\right)=\left(a^2+b^2+c^2\right)\) =1
=> \(\frac{ab+2c^2}{...}\ge\frac{ab+2c^2}{1}=2c^2+ab\)
tương tự + vào thì e sẽ ra điều phải chứng minh
Nhà hàng Tôm hùm kính chào quý khách ĐC : 255 Nguyễn Huệ, Q tân bình , TP HCM
Áp dụng BĐT Bunhiacopxky ta có:
\(\left(a^2+2c^2\right)\left(1+2\right)\ge\left(a+2c^2\right)\)
\(\Rightarrow\sqrt{a^2+2c^2}\ge\frac{a+2c}{3}\)
\(\Rightarrow\frac{\sqrt{a^2+2c^2}}{ac}\ge\frac{a+2c}{\sqrt{3ac}}=\frac{ab+2bc}{\sqrt{3abc}}\)
\(\Rightarrow\hept{\begin{cases}\frac{\sqrt{c^2+2b^2}}{bc}\ge\frac{ac+2ab}{\sqrt{3abc}}\\\frac{\sqrt{b^2+2a^2}}{ab}\ge\frac{bc+2ac}{\sqrt{abc}}\end{cases}}\)
Ta được BĐT:
\(VT\ge\frac{1}{3}.\frac{ab+2abc+ac+2ab+bc+2ac}{abc}=\frac{1}{3}.\frac{3\left(ab+bc+ac\right)}{abc}\)
\(=\frac{1}{\sqrt{3}}.\frac{3abc}{abc}=3\)
=> đpcm
P/S: Làm tắt vs đoạn này k^o chắc mấy :V
Repair đề \(\Sigma_{cyc}\frac{\sqrt{2a^2+b^2}}{ab}\ge3\sqrt{3}\).Because dấu '=' xảy ra khi \(a=b=c=3\)
Không use condition của đề bài :))
Ta co:
\(VT=\sqrt{\frac{a}{b}+\frac{a}{b}+\frac{b}{a}}+\sqrt{\frac{b}{c}+\frac{b}{c}+\frac{c}{b}}+\sqrt{\frac{c}{a}+\frac{c}{a}+\frac{a}{c}}\)
\(\Rightarrow VT\ge\sqrt{3\sqrt[3]{\frac{a}{b}}}+\sqrt{3\sqrt[3]{\frac{b}{c}}}+\sqrt{3\sqrt[3]{\frac{c}{a}}}\ge3\sqrt[3]{\sqrt{3\sqrt[3]{\frac{a}{b}}.\sqrt{3\sqrt[3]{\frac{b}{c}}.\sqrt{3\sqrt[3]{\frac{c}{a}}}}}}=3\sqrt{3}\)
equelity iff \(a=b=c=3\)