khó quá bạn ơi mình cần thêm thời gian để làm 

nhanh lên nhé

\(VT\ge\dfrac{1}{\left(a^2+1\right)-1}+\dfrac{1}{\left(b^2+1\right)-1}+\dfrac{1}{\left(c^2+1\right)-1}+4-\dfrac{4}{ab+1}+4-\dfrac{4}{bc+1}+4-\dfrac{4}{ca+1}\)

\(VT\ge\dfrac{1}{a^2}+\dfrac{1}{b^2}+\dfrac{1}{c^2}-\dfrac{4}{ab+1}-\dfrac{4}{bc+1}-\dfrac{4}{ca+1}+12\)

Mặt khác \(a;b;c\ge1\Rightarrow\left(a-1\right)\left(b-1\right)\ge0\Rightarrow ab+1\ge a+b\) (và tương tự...)

\(\Rightarrow VT\ge\dfrac{1}{ab}+\dfrac{1}{bc}+\dfrac{1}{ca}-\dfrac{4}{a+b}-\dfrac{4}{b+c}-\dfrac{4}{c+a}+12\)

\(VT\ge\dfrac{4}{\left(a+b\right)^2}+\dfrac{4}{\left(b+c\right)^2}+\dfrac{4}{\left(c+a\right)^2}-\dfrac{4}{a+b}-\dfrac{4}{b+c}-\dfrac{4}{c+a}+1+1+1+9\)

\(VT\ge\left(\dfrac{2}{a+b}-1\right)^2+\left(\dfrac{2}{b+c}-1\right)^2+\left(\dfrac{2}{c+a}-1\right)^2+9\ge9\)

a p dg côsi \(a\sqrt{b-1}=a.1.\sqrt{b-1}\le a.\dfrac{1+b-1}{2}=\dfrac{ab}{2}\)

ttuong tu \(b\sqrt{a-1}\le\dfrac{ab}{2}\)

nên vt\(\le ab\)

dau = xảy ra a=b=2

Bài 1:

Áp dụng BĐT Bunhiacopxky:

\((a^2+b^2+c^2+d^2)(1+1+1+1)\geq (a+b+c+d)^2\)

\(\Leftrightarrow a^2+b^2+c^2+d^2\geq \frac{(a+b+c+d)^2}{4}=\frac{2^2}{4}=1\) (đpcm)

Dấu "=" xay ra khi \(a=b=c=d=\frac{1}{2}\)

Bài 2:

Bạn xem lại đề:

Áp dụng BĐT Cô-si cho các số không âm ta có:

\(16a^4+1\geq 2\sqrt{16a^4.1}=8a^2\Rightarrow \frac{a^2}{1+16a^4}\leq \frac{a^2}{8a^2}=\frac{1}{8}(1)\)

\(b^4+1\geq 2\sqrt{b^4.1}=2b^2\Rightarrow \frac{b^2}{1+b^4}\leq \frac{b^2}{2b^2}=\frac{1}{2}(2)\)

Từ \((1);(2)\Rightarrow \frac{a^2}{1+16a^4}+\frac{b^2}{1+b^4}\leq \frac{1}{8}+\frac{1}{2}=\frac{5}{8}\) chứ không phải $\frac{1}{4}$

Nếu bạn muốn kết quả là $\frac{1}{4}$ thì cần thay $b^4$ bằng $16b^4$ và làm tương tự như trên.

Bài 3:

Ta có:

\(2x+3y+\frac{6}{x}+\frac{10}{y}=\frac{1}{2}(x+y)+(\frac{3}{2}x+\frac{6}{x})+(\frac{5}{2}y+\frac{10}{y})\)

Áp dụng BĐT Cô-si cho các số dương:

\(\frac{3}{2}x+\frac{6}{x}\geq 2\sqrt{\frac{3}{2}x.\frac{6}{x}}=6(1)\)

\(\frac{5}{2}y+\frac{10}{y}\geq 2\sqrt{\frac{5}{2}y.\frac{10}{y}}=10(2)\)

\(\frac{1}{2}(x+y)\geq \frac{1}{2}.4=2(3)\) do $x+y\geq 4$

Từ \((1);(2);(3)\Rightarrow 2x+3y+\frac{6}{x}+\frac{10}{y}\geq 6+10+2=18\)(đpcm)

Dấu "=" xảy ra khi $x=y=2$.

Ta có \(\sqrt{a+b}=\sqrt{a-1}+\sqrt{b-1}\Leftrightarrow a+b=a-1+2\sqrt{\left(a-1\right)\left(b-1\right)}+b-1\Leftrightarrow2=2\sqrt{\left(a-1\right)\left(b-1\right)}\Leftrightarrow\sqrt{\left(a-1\right)\left(b-1\right)}=1\Leftrightarrow\left(a-1\right)\left(b-1\right)=1\Leftrightarrow ab-a-b+1=1\Leftrightarrow a+b=ab\)Vậy nếu \(\sqrt{a+b}=\sqrt{a-1}+\sqrt{b-1}\) thì a+b=ab

\(\sqrt{a+b}=\sqrt{a-1}+\sqrt{b-1}\left(a\ge1;b\ge1\right)\\ \Leftrightarrow a+b=a-1+b-1+2\sqrt{\left(a-1\right)\left(b-1\right)}\\ \Leftrightarrow a+b=a+b-2+2\sqrt{\left(a-1\right)\left(b-1\right)}\\ \Leftrightarrow2=2\sqrt{\left(a-1\right)\left(b-1\right)}\\ \Leftrightarrow1=\sqrt{a-1}\sqrt{b-1}\\ \Leftrightarrow1=\left(a-1\right)\left(b-1\right)\\ \Leftrightarrow1=ab-a-b-1\\ \Leftrightarrow ab=a+b\)

Ta có: \(\frac{1+ab}{1+a^2}+\frac{1+ab}{1+b^2}=\left(1+ab\right)\left(\frac{1}{1+a^2}+\frac{1}{1+b^2}\right)\)

mà \(\frac{1}{1+a^2}+\frac{1}{1+b^2}\ge\frac{4}{2+a^2+b^2}\)( Áp dụng BĐT phụ \(\frac{1}{a}+\frac{1}{b}\ge\frac{4}{a+b}\))

Mặt khác: \(a^2+b^2\ge2ab\)

=> \(\frac{1}{1+a^2}+\frac{1}{1+b^2}\ge\frac{4}{2+2ab}=\frac{2}{1+ab}\)

=> \(\left(1+ab\right)\left(\frac{1}{1+a^2}+\frac{1}{1+b^2}\right)\ge\left(1+ab\right)\left(\frac{2}{1+ab}\right)=2\)(đpcm)