\(\Leftrightarrow\frac{1}{2}\left[\left(a-b\right)^2+\left(b-c\right)^2+\left(c-a\right)^2\right]\ge0\)(Đúng)

"=" khi a=b=c

Ta có BĐT \(\Leftrightarrow2a^2+2b^2+2c^2-2ab-2bc-2ca\ge0\)

\(\Leftrightarrow\left(a-b\right)^2+\left(b-c\right)^2+\left(c-a\right)^2\ge0\)

Vì \(\hept{\begin{cases}\left(a-b\right)^2\ge0\\\left(b-c\right)^2\ge0\\\left(c-a\right)^2\ge0\end{cases}\Rightarrow}\)BĐT luôn đúng

a² + b² + c² ≥ ab + bc + ca

<=> 2( a² + b² + c² ) ≥ 2( ab + bc + ca )

<=> 2a² + 2b² + 2c² ≥ 2ab + 2bc + 2ca

<=> 2a² + 2b² + 2c² - 2ab - 2bc - 2ca ≥ 0

<=> ( a² - 2ab + b² ) + ( b² - 2bc + c² ) + ( c² - 2ca + a² ) ≥ 0

<=> ( a - b )² + ( b - c )² + ( c - a )² ≥ 0 ( đúng )

Vậy bđt được chứng minh

Đẳng thức xảy ra <=> \(\hept{\begin{cases}a-b=0\\b-c=0\\c-a=0\end{cases}}\Leftrightarrow a=b=c\)

\(BDT\Leftrightarrow\sum\left[\dfrac{\left(a+b\right)^2}{c^2+ab}-2\right]\ge0\)\(\Leftrightarrow\sum\dfrac{a^2+b^2-2c^2}{c^2+ab}\ge0\)(*)

\(\Leftrightarrow\sum\left(\dfrac{a^2-c^2}{c^2+ab}+\dfrac{b^2-c^2}{c^2+ab}\right)\ge0\)

\(\Leftrightarrow\sum\left(c^2-a^2\right)\left(\dfrac{1}{a^2+bc}-\dfrac{1}{c^2+ab}\right)\ge0\)

\(\Leftrightarrow\sum\left(c-a\right)^2.\dfrac{\left(c+a\right)\left(c+a-b\right)}{\left(a^2+bc\right)\left(c^2+ab\right)}\ge0\)

\(\dfrac{\left(a+b\right)^2}{c^2+ab}+\dfrac{\left(b+c\right)^2}{a^2+bc}+\dfrac{\left(c+a\right)^2}{b^2+ca}\ge\dfrac{\left(a+b+b+c+c+a\right)^2}{a^2+b^2+c^2+ab+bc+ca}\)\(=\dfrac{4\left(a+b+c\right)^2}{a^2+b^2+c^2+ab+bc+ca}\) (theo AM-GM với a ; b>0)

\(=\dfrac{4\left(a^2+b^2+c^2+2ab+2bc+2ca\right)}{a^2+b^2+c^2+ab+bc+ca}=\dfrac{4.3.\left(a^2+b^2+c^2\right)}{2.\left(a^2+b^2+c^2\right)}\)(do \(a^2+b^2+c^2\ge ab+bc+ca\))

\(=4.1,5\) = 6 ( do a;b;c>0)

bđt \(\Leftrightarrow\)\(\Sigma_{cyc}\frac{a^2}{2}+\Sigma_{cyc}\frac{a}{bc}\ge\frac{9}{2}\)

mặt khác: \(\Sigma_{cyc}\frac{a}{bc}=\frac{1}{2}\Sigma_{cyc}\left(\frac{b}{ca}+\frac{c}{ab}\right)\ge\Sigma\frac{1}{a}\)\(\Rightarrow\)\(\Sigma_{cyc}\frac{a}{bc}\ge\Sigma_{cyc}\frac{1}{a}\)

do đó cần CM: \(\Sigma_{cyc}\frac{a^2}{2}+\Sigma_{cyc}\frac{1}{a}\ge\frac{9}{2}\) (1) 

\(VT_{\left(1\right)}=\Sigma_{cyc}\left(\frac{a^2}{2}+\frac{1}{2a}+\frac{1}{2a}\right)\ge3.\frac{3}{2}=\frac{9}{2}\)

"=" \(\Leftrightarrow\)\(a=b=c=1\)

Biến đổi tương đương:

\(\left(a+b+c\right)^2\ge3\left(ab+ac+bc\right)\)

\(\Leftrightarrow a^2+b^2+c^2+2ab+2ac+2bc\ge3\left(ab+ac+bc\right)\)

\(\Leftrightarrow a^2+b^2+c^2-ab-ac-bc\ge0\)

\(\Leftrightarrow2a^2+2b^2+2c^2-2ab-2ac-2bc\ge0\)

\(\Leftrightarrow\left(a-b\right)^2+\left(a-c\right)^2+\left(b-c\right)^2\ge0\) (luôn đúng)

Dấu "=" xảy ra khi \(a=b=c\)

\(\Rightarrow\frac{\left(a+b+c\right)^2}{ab+ac+bc}\ge3\)

b/ \(VT=\frac{\left(a+b+c\right)^2}{ab+ac+bc}+\frac{ab+ac+bc}{\left(a+b+c\right)^2}=\frac{8\left(a+b+c\right)^2}{9\left(ab+ac+bc\right)}+\frac{\left(a+b+c\right)^2}{9\left(ab+ac+bc\right)}+\frac{ab+ac+bc}{\left(a+b+c\right)^2}\)

\(\Rightarrow VT\ge\frac{8\left(a+b+c\right)^2}{9\left(ab+ac+bc\right)}+2\sqrt{\frac{\left(a+b+c\right)^2\left(ab+ac+bc\right)}{9\left(ab+ac+bc\right)\left(a+b+c\right)^2}}\ge\frac{8.3}{9}+\frac{2}{3}=\frac{10}{3}\) (đpcm)

Dấu "=" xảy ra khi \(a=b=c\)

B trước nhé:

Áp dụng bất đẳng thức cosi cho 2 số thực dương a^2 và b^2; b^2 và c^2 ; c^2 và a^2 ta được:

a^2 + b^2\(\ge\)2ab

Tương tự b^2 + c^2\(\ge\)2bc

Cx có c^2+a^2\(\ge\)2ac

=> 2(a^2+b^2+c^2)\(\ge\)2(ab + bc +ca)

=>a^2 + b^2 +c^2\(\ge\)ab+bc+ca

\(\dfrac{a^2+bc}{b+c}=\dfrac{\left(a+b\right)\left(a+c\right)-a\left(b+c\right)}{b+c}=\dfrac{\left(a+b\right)\left(a+c\right)}{b+c}-a\)

\(\Rightarrow VT=\dfrac{\left(a+b\right)\left(a+c\right)}{b+c}+\dfrac{\left(a+b\right)\left(b+c\right)}{a+c}+\dfrac{\left(a+c\right)\left(b+c\right)}{a+b}-\left(a+b+c\right)\)

Mặt khác áp dụng \(x+y+z\ge\sqrt{xy}+\sqrt{yz}+\sqrt{zx}\)

\(\Rightarrow\dfrac{\left(a+b\right)\left(a+c\right)}{b+c}+\dfrac{\left(a+b\right)\left(b+c\right)}{a+c}+\dfrac{\left(a+c\right)\left(b+c\right)}{a+b}\ge a+b+b+c+a+c=2\left(a+b+c\right)\)

\(\Rightarrow VT\ge2\left(a+b+c\right)-\left(a+b+c\right)=a+b+c\) (đpcm)

`a) 2 ( a^2 + b^2 ) >= ( a + b )^2`

`<=> 2a^2 + 2b^2 >= a^2 + 2ab + b^2`

`<=> a^2 - 2ab + b^2 >= 0`

`<=> ( a - b )^2 >= 0` (Luôn đúng `AA a,b`)

     `=>` Đẳng thức được c/m

_________________________________________

`b) a^2 + b^2 + c^2 >= ab + bc + ca`

`<=> 2a^2 + 2b^2 + 2c^2 >= 2ab + 2bc + 2ca`

`<=> ( a^2 - 2ab + b^2 ) + ( b^2 - 2bc + c^2 ) + ( c^2 - 2ca + a^2 ) >= 0`

`<=> ( a - b )^2 + ( b - c )^2 + ( c - a )^2 >= 0` (Luôn đúng `AA a,b,c`)

         `=>` Đẳng thức được c/m

\(2\left(a^2+b^2+c^2\right)\ge2\left(ab+bc+ca\right)\)

\(\Leftrightarrow3\left(a^2+b^2+c^2\right)\ge a^2+b^2+c^2+2\left(ab+bc+ca\right)\)

\(\Leftrightarrow a^2+b^2+c^2\ge\frac{\left(a+b+c\right)^2}{3}\)

Chuyển vế và CM tương tự

Bài 1 với bài 2 như nhau, đăng làm gì cho tốn công :))

Áp dụng bất đẳng thức Cauchy ta có :

\(\frac{ab}{c}+\frac{bc}{a}\ge2\sqrt{\frac{ab}{c}.\frac{bc}{a}}=2b\)

\(\frac{ab}{c}+\frac{ca}{b}\ge2\sqrt{\frac{ab}{c}.\frac{ca}{b}}=2a\)

\(\frac{ac}{b}+\frac{bc}{a}\ge2\sqrt{\frac{ac}{b}.\frac{bc}{a}}=2c\)

Cộng vế với vế ta được :

\(2\left(\frac{ab}{c}+\frac{bc}{a}+\frac{ca}{b}\right)\ge2\left(a+b+c\right)\)

\(\Rightarrow\frac{ab}{c}+\frac{bc}{a}+\frac{ca}{b}\ge a+b+c\)(đpcm)