Câu hỏi của Minh Thư

Question

Cho a,b,c thực dương t.m: a+b+c=2CMR: P = ab/căn ( ab+2c) + bc/căn( bc+2a) +ca/căn ( ca+2b)<=1

Kiệt Nguyễn · Accepted Answer

Ta có: a + b + c = 2 nên $2c+ab=c\left(a+b+c\right)+ab=ac+bc+c^2+ab$$=\left(ca+c^2\right)+\left(bc+ab\right)=c\left(a+c\right)+b\left(a+c\right)$$=\left(b+c\right)\left(a+c\right)$Áp dụng BĐT Cô - si cho 2 số không âm:$\frac{1}{b+c}+\frac{1}{a+c}\ge2\sqrt{\frac{1}{\left(b+c\right)\left(a+c\right)}}$(Vì a,b,c thực dương)$\Rightarrow\sqrt{\frac{1}{\left(b+c\right)\left(a+c\right)}}\le\frac{1}{2}\left(\frac{1}{b+c}+\frac{1}{a+c}\right)$$\Rightarrow\frac{1}{\sqrt{2c+ab}}\le\frac{1}{2}\left(\frac{1}{b+c}+\frac{1}{a+c}\right)$(cmt)$\Rightarrow\frac{ab}{\sqrt{ab+2c}}\le\frac{1}{2}\left(\frac{ab}{b+c}+\frac{ab}{a+c}\right)$(nhân 2 vế cho ab thực dương)    (1)(Dấu "="$\Leftrightarrow\frac{1}{b+c}=\frac{1}{c+a}\Leftrightarrow b+c=c+a\Leftrightarrow a=b$)Tương tự ta có: $\frac{bc}{\sqrt{bc+2a}}\le\frac{1}{2}\left(\frac{bc}{b+a}+\frac{bc}{a+c}\right)$(Dấu "="$\Leftrightarrow b=c$)  (2)$\frac{ca}{\sqrt{ca+2b}}\le\frac{1}{2}\left(\frac{ca}{c+b}+\frac{ca}{b+a}\right)$(Dấu "="$\Leftrightarrow a=c$)  (3)Cộng các BĐT (1) , (2) , (3), ta được:$P\le\frac{1}{2}\left(\frac{ab}{c+a}+\frac{ab}{c+b}+\frac{bc}{b+a}+\frac{cb}{c+a}+\frac{ac}{b+a}+\frac{ac}{c+b}\right)$$\Rightarrow P\le\frac{1}{2}\left(\frac{b\left(c+a\right)}{c+a}+\frac{a\left(c+b\right)}{c+b}+\frac{c\left(b+a\right)}{b+a}\right)$$\le\frac{1}{2}\left(a+b+c\right)=1$Vậy $P=\frac{ab}{\sqrt{ab+2c}}$$+\frac{bc}{\sqrt{bc+2a}}$$+\frac{ca}{\sqrt{ca+2b}}\le1$(Dấu "="$\Leftrightarrow a=b=c=\frac{2}{3}$)Câu trả lời: Ta có: a + b + c = 2 nên $2c+ab=c\left(a+b+c\right)+ab=ac+bc+c^2+ab$$=\left(ca+c^2\right)+\left(bc+ab\right)=c\left(a+c\right)+b\left(a+c\right)$$=\left(b+c\right)\left(a+c\right)$Áp dụng BĐT Cô - si cho 2 số không âm:$\frac{1}{b+c}+\frac{1}{a+c}\ge2\sqrt{\frac{1}{\left(b+c\right)\left(a+c\right)}}$(Vì a,b,c thực dương)$\Rightarrow\sqrt{\frac{1}{\left(b+c\right)\left(a+c\right)}}\le\frac{1}{2}\left(\frac{1}{b+c}+\frac{1}{a+c}\right)$$\Rightarrow\frac{1}{\sqrt{2c+ab}}\le\frac{1}{2}\left(\frac{1}{b+c}+\frac{1}{a+c}\right)$(cmt)$\Rightarrow\frac{ab}{\sqrt{ab+2c}}\le\frac{1}{2}\left(\frac{ab}{b+c}+\frac{ab}{a+c}\right)$(nhân 2 vế cho ab thực dương)    (1)(Dấu "="$\Leftrightarrow\frac{1}{b+c}=\frac{1}{c+a}\Leftrightarrow b+c=c+a\Leftrightarrow a=b$)Tương tự ta có: $\frac{bc}{\sqrt{bc+2a}}\le\frac{1}{2}\left(\frac{bc}{b+a}+\frac{bc}{a+c}\right)$(Dấu "="$\Leftrightarrow b=c$)  (2)$\frac{ca}{\sqrt{ca+2b}}\le\frac{1}{2}\left(\frac{ca}{c+b}+\frac{ca}{b+a}\right)$(Dấu "="$\Leftrightarrow a=c$)  (3)Cộng các BĐT (1) , (2) , (3), ta được:$P\le\frac{1}{2}\left(\frac{ab}{c+a}+\frac{ab}{c+b}+\frac{bc}{b+a}+\frac{cb}{c+a}+\frac{ac}{b+a}+\frac{ac}{c+b}\right)$$\Rightarrow P\le\frac{1}{2}\left(\frac{b\left(c+a\right)}{c+a}+\frac{a\left(c+b\right)}{c+b}+\frac{c\left(b+a\right)}{b+a}\right)$$\le\frac{1}{2}\left(a+b+c\right)=1$Vậy $P=\frac{ab}{\sqrt{ab+2c}}$$+\frac{bc}{\sqrt{bc+2a}}$$+\frac{ca}{\sqrt{ca+2b}}\le1$(Dấu "="$\Leftrightarrow a=b=c=\frac{2}{3}$)

coolkid · Answer

Ta có:$\frac{ab}{\sqrt{ab+2c}}=\frac{ab}{\sqrt{ab+\left(a+b+c\right)c}}=\frac{ab}{\sqrt{\left(c+a\right)\left(c+b\right)}}\le\frac{ab}{c+a}+\frac{ab}{c+b}$Tương tự:$\frac{bc}{\sqrt{bc+2a}}\le\frac{bc}{a+b}+\frac{bc}{a+c}$$\frac{ca}{\sqrt{ca+2b}}\le\frac{ca}{b+c}+\frac{ca}{b+a}$Khi đó:$P\le\frac{ab}{a+c}+\frac{ab}{c+b}+\frac{bc}{a+b}+\frac{bc}{a+c}+\frac{ca}{b+c}+\frac{ca}{b+a}$$=\frac{b\left(a+c\right)}{a+c}+\frac{a\left(b+c\right)}{b+c}+\frac{c\left(a+b\right)}{b+a}$$=a+b+c=2$Dấu "=" xảy ra tại $a=b=c=\frac{2}{3}$Câu trả lời: Ta có:$\frac{ab}{\sqrt{ab+2c}}=\frac{ab}{\sqrt{ab+\left(a+b+c\right)c}}=\frac{ab}{\sqrt{\left(c+a\right)\left(c+b\right)}}\le\frac{ab}{c+a}+\frac{ab}{c+b}$Tương tự:$\frac{bc}{\sqrt{bc+2a}}\le\frac{bc}{a+b}+\frac{bc}{a+c}$$\frac{ca}{\sqrt{ca+2b}}\le\frac{ca}{b+c}+\frac{ca}{b+a}$Khi đó:$P\le\frac{ab}{a+c}+\frac{ab}{c+b}+\frac{bc}{a+b}+\frac{bc}{a+c}+\frac{ca}{b+c}+\frac{ca}{b+a}$$=\frac{b\left(a+c\right)}{a+c}+\frac{a\left(b+c\right)}{b+c}+\frac{c\left(a+b\right)}{b+a}$$=a+b+c=2$Dấu "=" xảy ra tại $a=b=c=\frac{2}{3}$

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