Thêm điều kiện là a,b cùng dấu nha! mình đánh thiếu

1) Áp dụng bunhiacopxki ta được \(\sqrt{\left(2a^2+b^2\right)\left(2a^2+c^2\right)}\ge\sqrt{\left(2a^2+bc\right)^2}=2a^2+bc\), tương tự với các mẫu ta được vế trái \(\le\frac{a^2}{2a^2+bc}+\frac{b^2}{2b^2+ac}+\frac{c^2}{2c^2+ab}\le1< =>\)\(1-\frac{bc}{2a^2+bc}+1-\frac{ac}{2b^2+ac}+1-\frac{ab}{2c^2+ab}\le2< =>\)

\(\frac{bc}{2a^2+bc}+\frac{ac}{2b^2+ac}+\frac{ab}{2c^2+ab}\ge1\)<=> \(\frac{b^2c^2}{2a^2bc+b^2c^2}+\frac{a^2c^2}{2b^2ac+a^2c^2}+\frac{a^2b^2}{2c^2ab+a^2b^2}\ge1\)  (1) 

áp dụng (x² +y² +z²)(m²+n²+p²) \(\ge\left(xm+yn+zp\right)^2\)

(2a²bc +b²c² + 2b²ac+a²c² + 2c²ab+a²b²). VT\(\ge\left(bc+ca+ab\right)^2\)   <=> (ab+bc+ca)². VT \(\ge\left(ab+bc+ca\right)^2< =>VT\ge1\)  ( vậy (1) đúng)

dấu '=' khi a=b=c

4b, \(\frac{a^3}{a^2+b^2}+\frac{b^3}{b^2+c^2}+\frac{c^3}{c^2+a^2}=1-\frac{ab^2}{a^2+b^2}+1-\frac{bc^2}{b^2+c^2}+1-\frac{ca^2}{a^2+c^2}\)

\(\ge3-\frac{ab^2}{2ab}-\frac{bc^2}{2bc}-\frac{ca^2}{2ac}=3-\frac{\left(a+b+c\right)}{2}=\frac{3}{2}\)

4c, 

\(\frac{a+1}{b^2+1}+\frac{b+1}{c^2+1}+\frac{c+1}{a^2+1}=a+b+c-\frac{b^2}{b^2+1}-\frac{c^2}{c^2+1}-\frac{a^2}{a^2+1}+3--\frac{b^2}{b^2+1}-\frac{c^2}{c^2+1}-\frac{a^2}{a^2+1}\)\(\ge6-2\cdot\frac{\left(a+b+c\right)}{2}=3\)

                      Bài làm :

Ta có :

\(\left(a+b+c\right)^2=a^2+b^2+c^2\)

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

\(\Leftrightarrow2ab+2bc+2ac=0\)

\(\Leftrightarrow2\left(ab+bc+ac\right)=0\)

\(\Leftrightarrow ab+bc+ac=0\)

\(\Leftrightarrow\frac{ab+bc+ac}{abc}=0\)

\(\Leftrightarrow\frac{ab}{abc}+\frac{bc}{abc}+\frac{ac}{abc}=0\)

\(\Leftrightarrow\frac{1}{c}+\frac{1}{a}+\frac{1}{b}=0\)

\(\Leftrightarrow\frac{1}{a}+\frac{1}{b}=-\frac{1}{c}\left(1\right)\)

\(\Leftrightarrow\left(\frac{1}{a}+\frac{1}{b}\right)^3=\left(-\frac{1}{c}\right)^3\)

\(\Leftrightarrow\frac{1}{a^3}+\frac{1}{b^3}+\frac{3}{ab}\left(\frac{1}{a}+\frac{1}{b}\right)=-\frac{1}{c^3}\left(2\right)\)

Thay (1) vào (2) ; ta được :

\(\frac{1}{a^3}+\frac{1}{b^3}-\frac{3}{abc}=-\frac{1}{c^3}\)

\(\Leftrightarrow\frac{1}{a^3}+\frac{1}{b^3}+\frac{1}{c^3}=\frac{3}{abc}\)

=> Điều phải chứng minh

Ta có \(\left(a+b+c\right)^2=a^2+b^2+c^2\Leftrightarrow a^2+b^2+c^2+2ab+2ac+2bc=a^2+b^2+c^2\)

\(\Leftrightarrow2ab+2ac+2bc=0\)

\(\Leftrightarrow2\left(ab+ac+bc\right)=0\)

\(\Leftrightarrow ab+ac+bc=0\)

Ta lại có giả sử

\(\frac{1}{a^3}+\frac{1}{b^3}+\frac{1}{c^3}=\frac{3}{abc}\)

\(\Leftrightarrow\frac{a^3b^3+b^3c^3+c^3a^3}{a^3b^3c^3}=\frac{3}{abc}\)

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

\(\Leftrightarrow a^3b^3+b^3c^3+c^3a^3=3.a^2b^2c^2\)

\(\Leftrightarrow a^3b^3+b^3c^3+c^3a^3-3.a^2b^2c^2=0\)

\(\Leftrightarrow\left(ab+bc+ac\right)^3-3ca\left(ab+bc\right)\left(ab+bc+ac\right)-3ab^3c\left(-ac\right)-3a^2b^2c^2=0\)

\(\Leftrightarrow0+3a^2b^2c^2-3a^2b^2c^2+0=0\)

\(\Leftrightarrow0=0\left(lđ\right)\)

Vậy bất đẳng thức được chứng minh 

ta cm bài toán phụ nếu a+b+c=0 thì \(a^3+b^3+c^3=3abc\)

Thật vậy ta có \(a+b=-c\)=>(a+b)³=-c³<=> a³+b³+3ab(a+b)=-c³

=> a³+b³+c³=-3ab(-c)=3abc                (a+b=-c)

trở về bài toán ta có\(\left(a+b+c\right)^2=a^2+b^2+c^2\)=>\(ab+bc+ca=0\)

=>\(\frac{ab+bc+ca}{abc}=0\)=>\(\frac{1}{a}+\frac{1}{b}+\frac{1}{c}=0\)

ÁP DỤNG BÀI TOÁN PHỤ ta có \(\frac{1}{a}+\frac{1}{b}+\frac{1}{c}=0\) =>\(\frac{1}{a^3}+\frac{1}{b^3}+\frac{1}{c^3}=3\frac{1}{a}\frac{1}{b}\frac{1}{c}=\frac{3}{abc}\)

Ta có : \(\frac{a^2}{b^2}+\frac{b^2}{a^2}+4\ge3\left(\frac{a}{b}+\frac{b}{a}\right)\)(1) . Đặt \(x=\frac{a}{b}+\frac{b}{a}\)

\(\Rightarrow\left|x\right|=\left|\frac{a}{b}+\frac{b}{a}\right|=\left|\frac{a}{b}\right|+\left|\frac{b}{a}\right|\ge2\) \(\Rightarrow\orbr{\begin{cases}x\ge2\\x\le-2\end{cases}}\)

bpt (1) \(\Leftrightarrow\left(x^2-2\right)+4\ge3x\Leftrightarrow x^2-3x+2\ge0\)

Xét bất phương trình sau : \(y^2-3y+2\ge0\Leftrightarrow\left(y-1\right)\left(y-2\right)\ge0\Leftrightarrow\orbr{\begin{cases}y\ge2\\y\le1\end{cases}}\)

Từ \(\orbr{\begin{cases}x\ge2\\x\le-2\end{cases}}\) suy ra x nằm trong miền nghiệm của bất  phương trình đang xét , vậy x phải  thỏa mãn  \(y^2-3y+2\ge0\), tức là \(x^2-3x+2\ge0\)đúng.

Suy ra (1) đúng. Vậy ta có đpcm 

+TH1: a, b trái dấu \(\Rightarrow\frac{a}{b}+\frac{b}{a}\le0\)

\(\Rightarrow VT>0\ge VP\), bất đẳng thức luôn đúng

+TH2: a, b cùng dấu \(\Rightarrow\frac{a}{b}+\frac{b}{a}=\left|\frac{a}{b}\right|+\left|\frac{b}{a}\right|\ge2\sqrt{\left|\frac{a}{b}\right|.\left|\frac{b}{a}\right|}=2\)

bđt \(\Leftrightarrow\left(\frac{a}{b}+\frac{b}{a}\right)^2+2\ge3\left(\frac{a}{b}+\frac{b}{a}\right)\)

Đặt \(t=\frac{a}{b}+\frac{b}{a}\ge2\)

Cần chứng minh \(t^2+2\ge3t\Leftrightarrow\left(t-1\right)\left(t-2\right)\ge0\text{ }\left(\text{đúng }\forall t\ge2\right)\)

B1:

Từ \(b=\frac{a+c}{2}\Rightarrow2b=a+c\left(1\right)\)

Từ \(c=\frac{2bd}{b+a}\)thay vào (1) ta được:

\(2b=a+\frac{2bd}{b+a}\)

\(\Leftrightarrow2b\left(b+a\right)=a\left(b+a\right)+2bd\)

\(\Leftrightarrow2b^2+2ab=ab+a^2+2bd\)

\(\Leftrightarrow2b^2+ab-a^2-2bd=0\)

\(\Leftrightarrow2b\left(b-d\right)+a\left(b-a\right)=0\)

\(\Leftrightarrow2b\left(b-d\right)=a\left(a-b\right)\Leftrightarrow\frac{2b}{a}=\frac{a-b}{b-d}\)

B2: Từ \(\frac{1}{c}=\frac{1}{2}\left(\frac{1}{a}+\frac{1}{b}\right)\Rightarrow\frac{1}{c}=\frac{a+b}{2ab}hay2ab=c\left(a+b\right)\)

\(\Rightarrow ab+ab=ac+bc\Rightarrow ab-bc=ac-ab\Rightarrow b\left(a-c\right)=a\left(c-b\right)\)

Do đó: \(\frac{a-c}{c-b}=\frac{a}{b}\)(đpcm)