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\(VT=\sqrt{\left(a+b\right)\left(a+c\right)}+\sqrt{\left(b+c\right)\left(b+a\right)}+\sqrt{\left(c+a\right)\left(c+b\right)}\le_{AM-GM}\dfrac{a+b+a+c}{2}+\dfrac{b+c+b+a}{2}+\dfrac{c+a+c+b}{2}=2\left(a+b+c\right)=VP\) (đpcm)
Đầy đủ hơn 1 tí nhé
Theo gt : ab + bc + ca = 1 nên a2 + 1 = a2 + ab + bc + ca
= ( a + b )( a + c )
- Áp dụng bđt Cauchy ta có :
\(\sqrt{a^2+1}=\sqrt{\left(a+b\right)\left(a+c\right)}\le\frac{\left(a+b\right)\left(a+c\right)}{2}\)
- Tương tư ta cũng có :
\(\sqrt{b^2+1}\le\frac{\left(b+a\right)+\left(b+c\right)}{2}\)và \(\sqrt{c^2+1}\le\frac{\left(c+a\right)+\left(c+b\right)}{2}\)
Từ đó suy ra : VT \(\le\frac{\left(a+b\right)+\left(a+c\right)+\left(b+a\right)+\left(b+c\right)+\left(c+a\right)+\left(c+b\right)}{2}\)
\(\le2\left(a+b+c\right)=VP\left(đpcm\right)\)
\(\sqrt{a^2+1}=\sqrt{a^2+ab+ac+bc}=\sqrt{a\left(a+b\right)+c\left(a+b\right)}=\sqrt{\left(a+c\right)\left(a+b\right)}\)
Áp dụng BĐT Cauchy: \(\sqrt{\left(a+b\right)\left(a+c\right)}\le\dfrac{a+b+a+c}{2}=\dfrac{2a+b+c}{2}\)
\(\Rightarrow\sqrt{1+a^2}\le\dfrac{2a+b+c}{2}\)
Chứng minh tương tự ta được: \(\left\{{}\begin{matrix}\sqrt{1+b^2}\le\dfrac{2b+a+c}{2}\\\sqrt{1+c^2}\le\dfrac{2c+a+b}{2}\end{matrix}\right.\)
Cộng vế với vế ta được:
\(\sqrt{1+a^2}+\sqrt{1+b^2}+\sqrt{1+c^2}\le\dfrac{4\left(a+b+c\right)}{2}=2\left(a+b+c\right)\)
Dấu "=" xảy ra khi \(a=b=c=\dfrac{1}{\sqrt{3}}\)
Đẳng thức quen thuộc: \(a^2+ab+bc+ca=\left(a+b\right)\left(a+c\right)\) và tương tự cho các mẫu số còn lại
Ta có:
\(\sum\dfrac{1}{a^2+1}=\sum\dfrac{1}{\left(a+b\right)\left(a+c\right)}=\dfrac{2\left(a+b+c\right)}{\left(a+b\right)\left(b+c\right)\left(c+a\right)}=\dfrac{2\left(ab+bc+ca\right)\left(a+b+c\right)}{\left(a+b\right)\left(b+c\right)\left(c+a\right)}\)
Mặt khác:
\(2\left(ab+bc+ca\right)\left(a+b+c\right)=\left[a\left(b+c\right)+b\left(c+a\right)+c\left(a+b\right)\right]\left(a+b+c\right)\)
\(\ge\left(a\sqrt{b+c}+b\sqrt{c+a}+c\sqrt{a+b}\right)^2\) (Bunhiacopxki)
\(\Rightarrow\sum\dfrac{1}{a^2+1}\ge\dfrac{\left(a\sqrt{b+c}+b\sqrt{c+a}+c\sqrt{a+b}\right)^2}{\left(a+b\right)\left(b+c\right)\left(c+a\right)}\)
\(=\left(\dfrac{a}{\sqrt{\left(a+b\right)\left(b+c\right)}}+\dfrac{b}{\sqrt{\left(a+b\right)\left(b+c\right)}}+\dfrac{c}{\sqrt{\left(a+c\right)\left(b+c\right)}}\right)^2\)
\(=\left(\dfrac{a}{\sqrt{a^2+1}}+\dfrac{b}{\sqrt{b^2+1}}+\dfrac{c}{\sqrt{c^2+1}}\right)^2\)
Do đó ta chỉ cần chứng minh:
\(\dfrac{a}{\sqrt{a^2+1}}+\dfrac{b}{\sqrt{b^2+1}}+\dfrac{c}{\sqrt{c^2+1}}\le\dfrac{3}{2}\)
\(\Leftrightarrow\dfrac{a}{\sqrt{\left(a+b\right)\left(a+c\right)}}+\dfrac{b}{\sqrt{\left(a+b\right)\left(b+c\right)}}+\dfrac{c}{\sqrt{\left(a+c\right)\left(b+c\right)}}\le\dfrac{3}{2}\)
Đúng theo AM-GM:
\(\sum\dfrac{a}{\sqrt{\left(a+b\right)\left(a+c\right)}}\le\dfrac{1}{2}\sum\left(\dfrac{a}{a+b}+\dfrac{a}{a+c}\right)=\dfrac{3}{2}\)
Dấu "=" xảy ra khi \(a=b=c=\dfrac{1}{\sqrt{3}}\)
\(VT=\frac{ab+bc+ca}{ab}+\frac{ab+bc+ca}{bc}+\frac{ab+bc+ca}{ca}\)
\(=3+\frac{c\left(a+b\right)}{ab}+\frac{a\left(b+c\right)}{bc}+\frac{b\left(c+a\right)}{ca}\)(1)
Theo BĐT AM-GM: \(\frac{1}{2}\left[\frac{c\left(a+b\right)}{ab}+\frac{a\left(b+c\right)}{bc}\right]\ge\sqrt{\frac{\left(a+b\right)\left(b+c\right)}{b^2}}\)
Tương tự: \(\frac{1}{2}\left[\frac{a\left(b+c\right)}{bc}+\frac{b\left(c+a\right)}{ca}\right]\ge\sqrt{\frac{\left(a+c\right)\left(b+c\right)}{c^2}}\)
\(\frac{1}{2}\left[\frac{c\left(a+b\right)}{ab}+\frac{b\left(c+a\right)}{ca}\right]\ge\sqrt{\frac{\left(a+c\right)\left(a+b\right)}{a^2}}\)
Cộng theo vế 3 BĐT trên rồi thay vào 1 ta sẽ thu được đpcm.
Ta có : \(\sqrt{1+a^2}+\sqrt{1+b^2}+\sqrt{1+c^2}=\sqrt{ab+bc+ac+a^2}+\sqrt{ab+bc+ac+b^2}+\sqrt{ab+bc+ac+c^2}=\sqrt{\left(b+a\right)\left(a+c\right)}+\sqrt{\left(a+b\right)\left(b+c\right)}+\sqrt{\left(a+c\right)\left(c+b\right)}\)
\(\le\frac{a+c+b+c}{2}+\frac{a+b+b+c}{2}+\frac{a+c+a+b}{2}=2\left(a+b+c\right)\)
( áp dụng BĐT Cô - si cho các số a ; b ; c dương )
Dấu " = " xảy ra \(\Leftrightarrow\left\{{}\begin{matrix}ab+bc+ac=1\\a+c=b+c=a+b\end{matrix}\right.\)
\(\Leftrightarrow a=b=c=\frac{1}{\sqrt{3}}\)
Vậy ...
\(\left(a+b\right)\left(b+c\right)\left(c+a\right)+abc\)
\(=abc+a^2b+ab^2+a^2c+ac^2+b^2c+bc^2+abc+abc\)
\(=\left(a+b+c\right)\left(ab+bc+ca\right)\)( phân tích nhân tử các kiểu )
\(\Rightarrow\left(a+b\right)\left(b+c\right)\left(c+a\right)\ge\left(a+b+c\right)\left(ab+bc+ca\right)-abc\left(1\right)\)
\(a+b+c\ge3\sqrt[3]{abc};ab+bc+ca\ge3\sqrt[3]{a^2b^2c^2}\Rightarrow\left(a+b+c\right)\left(ab+bc+ca\right)\ge9abc\)
\(\Rightarrow-abc\ge\frac{-\left(a+b+c\right)\left(ab+bc+ca\right)}{9}\)
Khi đó:\(\left(a+b+c\right)\left(ab+bc+ca\right)-abc\)
\(\ge\left(a+b+c\right)\left(ab+bc+ca\right)-\frac{\left(a+b+c\right)\left(ab+bc+ca\right)}{9}\)
\(=\frac{8\left(a+b+c\right)\left(ab+bc+ca\right)}{9}\left(2\right)\)
Từ ( 1 ) và ( 2 ) có đpcm
gt <=> \(\frac{1}{ab}+\frac{1}{bc}+\frac{1}{ca}=1\)
Đặt: \(\frac{1}{a}=x;\frac{1}{b}=y;\frac{1}{c}=z\)
=> Thay vào thì \(VT=\frac{\frac{1}{xy}}{\frac{1}{z}\left(1+\frac{1}{xy}\right)}+\frac{1}{\frac{yz}{\frac{1}{x}\left(1+\frac{1}{yz}\right)}}+\frac{1}{\frac{zx}{\frac{1}{y}\left(1+\frac{1}{zx}\right)}}\)
\(VT=\frac{z}{xy+1}+\frac{x}{yz+1}+\frac{y}{zx+1}=\frac{x^2}{xyz+x}+\frac{y^2}{xyz+y}+\frac{z^2}{xyz+z}\ge\frac{\left(x+y+z\right)^2}{x+y+z+3xyz}\)
Có BĐT x, y, z > 0 thì \(\left(x+y+z\right)\left(xy+yz+zx\right)\ge9xyz\)Ta thay \(xy+yz+zx=1\)vào
=> \(x+y+z\ge9xyz=>\frac{x+y+z}{3}\ge3xyz\)
=> Từ đây thì \(VT\ge\frac{\left(x+y+z\right)^2}{x+y+z+\frac{x+y+z}{3}}=\frac{3}{4}\left(x+y+z\right)\ge\frac{3}{4}.\sqrt{3\left(xy+yz+zx\right)}=\frac{3}{4}.\sqrt{3}=\frac{3\sqrt{3}}{4}\)
=> Ta có ĐPCM . "=" xảy ra <=> x=y=z <=> \(a=b=c=\sqrt{3}\)
Gọi 1/4 số a là 0,25 . Ta có :
a . 3 - a . 0,25 = 147,07
a . (3 - 0,25) = 147,07 ( 1 số nhân 1 hiệu )
a . 2,75 = 147,07
a = 147,07 : 2,75
a = 53,48
mình nha
Áp dụng BĐT AM-GM ta có:
\(\sqrt{a^2+1}=\sqrt{a^2+ab+bc+ca}=\sqrt{\left(a+b\right)\left(a+c\right)}\)
\(\le\frac{a+b+a+c}{2}=\frac{2a+b+c}{2}\)
Tương tự cho 2 BĐT còn lại cũng có:
\(\sqrt{b^2+1}\le\frac{2b+c+a}{2};\sqrt{c^2+1}\le\frac{2c+a+b}{2}\)
Cộng theo vế 2 BĐT trên thu đc ĐPCM