a. Do vai trò của a;b;c là như nhau, không mất tính tổng quát giả sử \(a\ge b\ge c\)

BĐT tương đương:

\(\left(a-b\right)\left[a^2-ac+bc-b^2\right]+c\left(a-b\right)\left(b-c\right)\ge0\)

\(\Leftrightarrow\left(a-b\right)\left[\left(a-b\right)\left(a+b\right)-c\left(a-b\right)\right]+c\left(a-b\right)\left(b-c\right)\ge0\)

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

b.

Ta có: \(a^6+a^6+a^6+a^6+a^6+b^6\ge6\sqrt[6]{a^{30}b^6}=6a^5b\)

Tương tự: \(5b^6+c^6\ge6b^5c\) ; \(5c^6+a^6\ge6c^5a\)

Cộng vế với vế:

\(6\left(a^6+b^6+c^6\right)\ge6\left(a^5b+b^5c+c^5a\right)\)

bạn ơi, bài này sai đề rồi

Ta có: BĐT\(\Leftrightarrow\dfrac{a}{a+b}-\dfrac{1}{2}+\dfrac{b}{b+c}-\dfrac{1}{2}+\dfrac{c}{c+a}-\dfrac{1}{2}\ge0\)

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

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

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

\(\Leftrightarrow\dfrac{a-b}{2}\left(\dfrac{1}{a+b}-\dfrac{1}{b+c}\right)+\dfrac{a-c}{2}\left(\dfrac{1}{b+c}-\dfrac{1}{c+a}\right)\ge0\)

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

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

Vậy BĐT luôn đúng với \(a\ge b\ge c>0\)

tham khảo tại đây-_-

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Từ "Siêu tốc thần sầu" đến "tập thể dục" tha hồ luyện

!!!

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For \(a\geq b\geq c>0\) we obtain:

\(\sum_{cyc}\frac{a}{a+b}-\frac{3}{2}=\sum_{cyc}\left(\frac{a}{a+b}-\frac{1}{2}\right)=\sum_{cyc}\frac{a-b}{2(a+b)}\)

\(=\sum_{cyc}\frac{(a-b)(c^2+ab+ac+bc)}{2\prod\limits_{cyc}(a+b)}=\sum_{cyc}\frac{c^2a-c^2b}{2\prod\limits_{cyc}(a+b)}\)

\(=\sum_{cyc}\frac{a^2b-a^2c}{2\prod\limits_{cyc}(a+b)}=\frac{(a-b)(a-c)(b-c)}{2\prod\limits_{cyc}(a+b)}\geq0\)

hơn 1 năm rồi không ai làm :'(

a) Áp dụng bđt Cauchy ta có :

\(a+b\ge2\sqrt{ab}\)(1)

\(b+c\ge2\sqrt{bc}\)(2)

\(c+a\ge2\sqrt{ca}\)(3)

Nhân (1), (2), (3) theo vế

=> \(\left(a+b\right)\left(b+c\right)\left(c+a\right)\ge8\sqrt{a^2b^2c^2}=8\sqrt{\left(abc\right)^2}=8\left|abc\right|=8abc\)

=> đpcm

Dấu "=" xảy ra <=> a=b=c

b) Áp dụng bđt AM-GM ta có :

\(\frac{bc}{a}+\frac{ca}{b}\ge2\sqrt{\frac{bc}{a}\cdot\frac{ca}{b}}=2\sqrt{c^2}=2c\)

TT : \(\frac{ca}{b}+\frac{ab}{c}\ge2a\); \(\frac{bc}{a}+\frac{ab}{c}\ge2b\)

Cộng vế với vế

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

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

Dấu "=" xảy ra <=> a=b=c

\(\frac{a}{a+b}\)>=  \(\frac{a}{a+a}\)= \(\frac{1}{2}\)( vì a + a >= a + b vì a >= b ) 

\(\frac{b}{b+c}\) >= \(\frac{b}{b+b}\)= \(\frac{1}{2}\)( vì b + b >= b + c vì b >= c )

\(\frac{c}{c+a}\)>= \(\frac{c}{c+c}\)  = \(\frac{1}{2}\)( vì c + c >= c + a vì c>=0 )

Từ 3 điều này suy ra

\(\frac{a}{a+b}\)+ \(\frac{b}{b+c}\)+ \(\frac{c}{c+a}\)>=  \(\frac{3}{2}\)

dễ dàng c/m (x+y+z)(1/x+1/y+1/z) \(\ge\) 9,dấu "=" khi x=y=z (*)

a/a+b +b/b+c +c/c+a >= 3/2

<=>(a/b+c + 1) + (b/c+a + 1) + (c/a+b + 1) >= 3/2+1+1+1

<=>(a+b+c)/(b+c) + (a+b+c)/(c+a) + (a+b+c)/(a+b) >= 9/2

<=>2(a+b+c)(1/b+c + 1/c+a + 1/a+b) >= 9/2

<=>[(b+c)+(c+a)+(a+b)](1/b+c + 1/c+a + 1/a+b) >= 9/2 (bđt (*))

Đặt: a + b = x; b + c = y; c + a = z

Thì ta có: x \(\ge\)z \(\ge\)y

Theo đề bài ta có:

\(\frac{a}{a+b}+\frac{b}{b+c}+\frac{c}{c+a}\ge\frac{3}{2}\)

\(\Leftrightarrow\frac{a}{a+b}-\frac{1}{2}+\frac{b}{b+c}-\frac{1}{2}+\frac{c}{c+a}-\frac{1}{2}\ge0\)

\(\Leftrightarrow\frac{a-b}{2\left(a+b\right)}+\frac{b-c}{2\left(b+c\right)}+\frac{c-a}{2\left(c+a\right)}\ge0\)

\(\Leftrightarrow\frac{z-y}{2x}+\frac{x-z}{2y}+\frac{y-x}{2z}\ge0\)

\(\Leftrightarrow xy^2+yz^2+zx^2-x^2y-y^2z-z^2x\ge0\)

\(\Leftrightarrow\left(y-x\right)\left(z-y\right)\left(z-x\right)\ge0\)(1)

Mà ta lại có 

\(\hept{\begin{cases}y-x\le0\\z-x\le0\\z-y\ge0\end{cases}}\)nên (1) đúng

\(\Rightarrow\)ĐPCM

Đấu = xảy ra khi x = y = z hay a = b = c

Đặt b+c=m

      a+c=n

      a+b=p

=>a+b+c =\(\frac{m+n+p}{2}\) 

a=\(\frac{n+p-m}{2}\) 

b=\(\frac{m+p-n}{2}\) 

c=\(\frac{m+n-p}{2}\) 

=>\(\frac{n+p-m}{2m}+\frac{m+n-p}{2n}+\frac{m+n-p}{2p}\) 

=\(\frac{1}{2}\left(\frac{n}{m}+\frac{m}{n}\right)\) +\(\frac{1}{2}\left(\frac{p}{m}+\frac{m}{p}\right)\) +\(\frac{1}{2}\left(\frac{p}{n}+\frac{n}{p}\right)\) -\(\frac{3}{2}\) \(\ge\) \(\frac{3}{2}\) 

Áp dụng BĐT Cosi cho  2 số \(\frac{n}{m};\frac{m}{n}\)  ta được:

 Từ chứng minh tiếp ....

Đã làm, làm tới cùng xem nào? 

Ta có: \(\dfrac{a}{a+b}+\dfrac{b}{b+c}+\dfrac{c}{c+a}\ge\dfrac{a}{2b}+\dfrac{b}{2c}+\dfrac{c}{2a}=\dfrac{1}{2}\left(\dfrac{a}{b}+\dfrac{b}{c}+\dfrac{c}{a}\right)\)

\(\ge\dfrac{1}{2}.3=\dfrac{3}{2}\) ( BĐT AM - GM )

Dấu " = " khi a = b = c

\(\Rightarrowđpcm\)

BĐT\(\Leftrightarrow\dfrac{a}{a+b}-\dfrac{1}{2}+\dfrac{b}{b+c}-\dfrac{1}{2}+\dfrac{c}{c+a}-\dfrac{1}{2}\ge0\)

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

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

\(\Leftrightarrow\dfrac{a-b}{2}\left(\dfrac{1}{a+b}-\dfrac{1}{b+c}\right)+\dfrac{a-c}{2}\left(\dfrac{1}{b+c}-\dfrac{1}{c+a}\right)\ge0\)

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

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

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

\(\Leftrightarrow\dfrac{\left(a-b\right)\left(a-c\right)\left(b-c\right)}{2\left(a+b\right)\left(b+c\right)\left(c+a\right)}\ge0\)(luôn đúng)

\(\Rightarrowđpcm\)

Áp dụng BĐT Cauchy dạng engel ta có:

\(\frac{a^2}{b}+\frac{b^2}{c}+\frac{c^2}{a}\ge\frac{(a+b+c)^2}{a+b+c}=a+b+c(đpcm) \)

theo bđt cauchy ta có

\(\left\{{}\begin{matrix}\dfrac{a^2}{b}+b\ge2a\\\dfrac{b^2}{c}+c\ge2b\\\dfrac{c^2}{a}+a\ge2c\end{matrix}\right.\)

\(\Leftrightarrow\dfrac{a^2}{b}+\dfrac{b^2}{c}+\dfrac{c^2}{a}+a+b+c\ge2a+2b+2c\)

\(\Leftrightarrow\dfrac{a^2}{b}+\dfrac{b^2}{c}+\dfrac{c^2}{a}\ge a+b+c\)

\(\Rightarrow dpcm\)