cậu thử biến đổi tương xem thế nào....

khó thế

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

\(\Leftrightarrow\)\(a^2b^2c^4=1-2abc\left(a+b\right)+a^2b^2c^2\left(a+b\right)^2\)

\(VT=\frac{\left(1+b^2c^2\right)\left(1+a^2c^2\right)}{c^2+a^2b^2c^2}=\frac{1+a^2c^2+b^2c^2+a^2b^2c^4}{c^2+a^2b^2c^2}\)

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

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

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

\(=\frac{2-2abc.\frac{1}{abc}}{c^2+a^2b^2c^2}+\left(a+b\right)^2=\left(a+b\right)^2=VP\) ( đpcm ) 

PS : sorry for late :'< 

Bài 1:

$\frac{a}{b}=\frac{c}{d}=t\Rightarrow a=bt; c=dt$. Khi đó:

\(\frac{2a^2-3ab+5b^2}{2a^2+3ab}=\frac{2(bt)^2-3.bt.b+5b^2}{2(bt)^2+3bt.b}=\frac{b^2(2t^2-3t+5)}{b^2(2t^2+3t)}\)

$=\frac{2t^2-3t+5}{2t^2+3t}(1)$
\(\frac{2c^2-3cd+5d^2}{2c^2+3cd}=\frac{2(dt)^2-3.dt.d+5d^2}{2(dt)^2+3dt.d}=\frac{d^2(2t^2-3t+5)}{d^2(2t^2+3t)}=\frac{2t^2-3t+5}{2t^2+3t}(2)\)

Từ $(1);(2)$ suy ra đpcm.

Bài 2:

Từ $\frac{a}{c}=\frac{c}{b}\Rightarrow c^2=ab$. Khi đó:

$\frac{b^2-c^2}{a^2+c^2}=\frac{b^2-ab}{a^2+ab}=\frac{b(b-a)}{a(a+b)}$ (đpcm)

Bài 3:

Đặt $\frac{a}{b}=\frac{c}{d}=t\Rightarrow a=bt; c=dt$

Khi đó:

$\frac{3a^6+c^6}{3b^6+d^6}=\frac{3(bt)^6+(dt)^6}{3b^6+d^6}=\frac{t^6(3b^6+d^6)}{3b^6+d^6}=t^6(*)$

Và:

$\frac{(a+c)^6}{(b+d)^6}=(\frac{bt+dt}{b+d})^6=t^6(**)$

Từ $(*); (**)\Rightarrow $ đpcm.

Ta có:

sigma \(\frac{ab}{3a+4b+5c}=\) sigma \(\frac{2ab}{5\left(a+b+2c\right)+\left(a+3b\right)}\le\frac{2}{36}\left(sigma\frac{5ab}{a+b+2c}+sigma\frac{ab}{a+3b}\right)\)

Ta đi chứng minh: \(sigma\frac{ab}{a+b+2c}\le\frac{9}{4}\)

có: \(sigma\frac{ab}{a+b+2c}\le\frac{1}{4}\left(sigma\frac{ab}{c+a}+sigma\frac{ab}{b+c}\right)=\frac{1}{4}\left(a+b+c\right)=\frac{9}{4}\)

BĐT trên đúng nếu: \(sigma\frac{ab}{a+3b}\le\frac{9}{4}\)

Ta thấy: \(sigma\frac{ab}{a+3b}\le\frac{1}{16}\left(sigma\frac{ab}{a}+sigma\frac{3ab}{b}\right)=\frac{1}{16}\)( sigma \(b+sigma3a\)) \(=\frac{1}{4}\left(a+b+c\right)=\frac{9}{4}\)

\(\Leftrightarrow sigma\frac{ab}{3a+4b+5c}\le\frac{1}{18}\left(5.\frac{9}{4}+\frac{9}{4}\right)=\frac{3}{4}\)(1)

MÀ: \(\frac{1}{\sqrt{ab\left(a+2c\right)\left(b+2c\right)}}=\frac{2}{2\sqrt{\left(ab+2bc\right)\left(ab+2ca\right)}}\ge\frac{2}{2\left(ab+bc+ca\right)}\)

\(=\frac{3}{3\left(ab+bc+ca\right)}\ge\frac{3}{\left(a+b+c\right)^2}=\frac{3}{9^2}=\frac{1}{27}\)(2)

Từ (1) và (2) \(\Rightarrow T\le\frac{3}{4}-\frac{1}{27}=\frac{77}{108}\)

Vậy GTLN của biểu thức T là 77/108 <=> a=b=c=3

\(\frac{a}{a+b}+\frac{b}{b+c}+\frac{c}{c+d}+\frac{d}{d+a}=2\Leftrightarrow1-\frac{a}{a+b}-\frac{b}{b+c}+1-\frac{c}{c+d}-\frac{d}{d+a}=0\)

\(\Leftrightarrow\frac{b\left(c-a\right)}{\left(a+b\right)\left(b+c\right)}+\frac{d\left(a-c\right)}{\left(c+d\right)\left(d+a\right)}=0\)

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

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

\(\Leftrightarrow bad+bd^2+bca+bcd-dab-dac-db^2-cbd=0\)

\(\Leftrightarrow bca-dca+bd^2-db^2=0\)

\(\Leftrightarrow\left(b-d\right)\left(ca-bd\right)=0\)

\(\Rightarrow ca=bd\Rightarrow abcd=bd^2\)