Ta có :

 \(\left(a+b+c\right)^2=a^2+b^2+c^2\)

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

\(\Rightarrow2\left(ab+bc+ca\right)=0\)

\(\Rightarrow ab+bc+ca=0\)

\(\Rightarrow\frac{ab+bc+ca}{abc}=0\)

\(\Rightarrow\frac{ab}{abc}+\frac{bc}{abc}+\frac{ca}{abc}=0\)

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

\(\Rightarrow\frac{1}{a}+\frac{1}{b}=-\frac{1}{c}\)

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

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

\(\Rightarrow\frac{1}{a^3}+\frac{1}{b^3}+\frac{1}{c^3}+\frac{3}{ab\left(-\frac{1}{c}\right)}=0\)

\(\Rightarrow\frac{1}{a^3}+\frac{1}{b^3}+\frac{1}{c^3}-\frac{3}{abc}=0\)

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

\(\frac{a^2}{b^2}+\frac{b^2}{c^2}\ge2\sqrt{\frac{a^2}{b^2}.\frac{b^2}{c^2}}=2\frac{a}{c}\\ \frac{a^2}{b^2}+\frac{c^2}{a^2}\ge2\frac{c}{b}\\ \frac{b^2}{c^2}+\frac{c^2}{a^2}\ge\frac{2b}{a}\)

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

=> đpcm

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, 

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

Giải:
Đặt \(\frac{a}{b}=\frac{c}{d}=k\)

\(\Rightarrow a=bk,c=dk\)

Ta có: \(\frac{a.b}{c.d}=\frac{bkb}{dkd}=\frac{b^2}{d^2}\) (1)

\(\frac{\left(a+b\right)^2}{\left(c+d\right)^2}=\frac{\left(bk+b\right)^2}{\left(dk+d\right)^2}=\frac{\left[b\left(k+1\right)\right]^2}{\left[d\left(k+1\right)\right]^2}=\frac{b^2}{d^2}\) (2)

\(\frac{a^2+b^2}{c^2+d^2}=\frac{\left(bk\right)^2+b^2}{\left(dk\right)^2+d^2}=\frac{b^2.k^2+b^2}{d^2.k^2+d^2}=\frac{b^2.\left(k^2+1\right)}{d^2.\left(k^2+1\right)}=\frac{b^2}{d^2}\) (3)

Từ (1), (2) và (3) suy ra \(\frac{a.b}{c.d}=\frac{\left(a+b\right)^2}{\left(c+d\right)^2}=\frac{a^2+b^2}{c^2+d^2}\)

Đặt \(\frac{a}{b}=\frac{c}{d}=k\Rightarrow\begin{cases}a=bk\\c=dk\end{cases}\)

ta có: \(\frac{a.b}{c.d}=\frac{b^2.k}{d^2.k}=\frac{b^2}{d^2}\left(1\right)\)

\(\frac{\left(a+b\right)^2}{\left(c+d\right)^2}=\frac{a^2+2ab+b^2}{c^2+2cd+d^2}=\frac{b^2.k^2+2b^2.k+b^2}{d^2.k^2+2d^2.k+d^2}=\frac{b^2}{d^2}\left(2\right)\)

\(\frac{a^2+b^2}{c^2+d^2}=\frac{b^2.k^2+b^2}{d^2.k^2+d^2}=\frac{b^2}{d^2}\left(3\right)\)

từ 1,2 và 3 ta có điều phải chứng minh

Ta có: \(\left(\frac{a}{b+c}+\frac{b}{a+c}+\frac{c}{a+b}\right)\left(a+b+c\right)=1.\left(a+b+c\right)\)

=>\(\frac{a^2}{b+c}+\frac{a\left(b+c\right)}{b+c}+\frac{b^2}{a+c}+\frac{b\left(a+c\right)}{a+c}+\frac{c^2}{a+b}+\frac{c\left(a+b\right)}{a+b}=a+b+c\)

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

=> \(\frac{a^2}{b+c}+\frac{b^2}{a+c}+\frac{c^2}{a+b}=0\)

Ta có b² = ac

=> b.b = a.c

=> \(\frac{a}{b}=\frac{b}{c}\)

Đặt \(\frac{a}{b}=\frac{b}{c}=k\Rightarrow\hept{\begin{cases}a=bk\\b=ck\end{cases}}\)

Khi đó \(\frac{a^2+b^2}{b^2+c^2}=\frac{\left(bk\right)^2+b^2}{\left(ck\right)^2+c^2}=\frac{b^2k^2+b^2}{c^2k^2+c^2}=\frac{b^2\left(k^2+1\right)}{c^2\left(k^2+1\right)}=\frac{b^2}{c^2}=\frac{\left(ck\right)^2}{c^2}=k^2\)(1)

; \(\frac{a}{c}=\frac{bk}{c}=\frac{ckk}{c}=k^2\)(2)

Từ (1)(2) => \(\frac{a^2+b^2}{b^2+c^2}=\frac{a}{c}\)

 \(\frac{a^2+b^2}{b^2+c^2}=\frac{a^2+ac}{ac+c^2}=\frac{a\left(a+c\right)}{c\left(a+c\right)}=\frac{a}{c}\)

=> đpcm

Đặt \(\frac{a}{b}=\frac{c}{d}=k\)

\(\Rightarrow a=bk;c=dk\)

Ta có:

\(\frac{7a^2+3ab}{11a^2-8b^2}=\frac{7b^2k^2+3\cdot bk\cdot b}{11b^2k^2-8b^2}=\frac{b^2\left(7k^2+3k\right)}{b^2\left(11k^2-8\right)}=\frac{7k^2+3k}{11k^2-8}\left(1\right)\)

\(\frac{7c^2+3cd}{11c^2-8d^2}=\frac{7d^2k^2+3dk\cdot d}{11d^2k^2-8d^2}=\frac{d^2\left(7k^2+3k\right)}{d^2\left(11k^2-8\right)}=\frac{7k^2+3k}{11k^2-8}\left(2\right)\)

Từ \(\left(1\right);\left(2\right)\Rightarrowđpcm\)

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