Cái c là \(\dfrac{2}{\sqrt{1+c^2}}\) ạ

\(P=\dfrac{2-\left(1+a^2\right)}{1+a^2}+\dfrac{2-\left(1+b^2\right)}{1+b^2}+\dfrac{2}{\sqrt{1+c^2}}\)

\(P=2\left(\dfrac{1}{1+a^2}+\dfrac{1}{1+b^2}+\dfrac{1}{\sqrt{1+c^2}}\right)-2\) 

Từ điều kiện \(ab+bc+ca=1\), đặt \(\left\{{}\begin{matrix}a=tanx\\b=tany\\c=tanz\end{matrix}\right.\) với \(x+y+z=\dfrac{\pi}{2}\)

Xét \(Q=\dfrac{1}{1+a^2}+\dfrac{1}{1+b^2}+\dfrac{1}{\sqrt{1+c^2}}=\dfrac{1}{1+tan^2x}+\dfrac{1}{1+tan^2y}+\dfrac{1}{\sqrt{1+tan^2z}}\)

\(Q=cos^2x+cos^2y+cosz=1+\dfrac{1}{2}\left(cos2x+cos2y\right)+cosz\)

\(=1+cos\left(x+y\right)cos\left(x-y\right)+cosz\le1+cos\left(x+y\right)+cosz\)

\(=1+cos\left(\dfrac{\pi}{2}-z\right)+cosz=1+sinz+cosz=1+\sqrt{2}sin\left(z+\dfrac{\pi}{4}\right)\le1+\sqrt{2}\)

\(\Rightarrow P\le2\left(1+\sqrt{2}\right)-2=2\sqrt{2}\)

Dấu "=" xảy ra khi \(\left\{{}\begin{matrix}x=y=\dfrac{\pi}{8}\\z=\dfrac{\pi}{4}\end{matrix}\right.\) \(\Rightarrow\left(a;b;c\right)=\left(\sqrt{2}-1;\sqrt{2}-1;1\right)\)

\(\dfrac{1}{\sqrt{a^2-ab+b^2}}< =\dfrac{1}{\sqrt{2ab-ab}}=\dfrac{1}{\sqrt{ab}}\)

\(\sqrt{\dfrac{1}{b^2-bc+c^2}}< =\dfrac{1}{\sqrt{bc}};\sqrt{\dfrac{1}{c^2-ac+c^2}}< =\dfrac{1}{\sqrt{ac}}\)

=>P<=1/a+1/b+1/c=3

Dấu = xảy ra khi a=b=c=1

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

Đặt \(\frac{1}{a}=x;\frac{1}{b}=y;\frac{1}{c}=z\)

\(\frac{a}{bc\left(a+1\right)}=\frac{\frac{1}{x}}{\frac{1}{y}\cdot\frac{1}{z}\left(\frac{1}{x}+1\right)}=\frac{xyz}{x\left(x+1\right)}=\frac{yz}{x+1}\)

Tươn tự rồi cộng vế theo vế:

\(A=\frac{xy}{z+1}+\frac{yz}{x+1}+\frac{zx}{y+1}\le\frac{\left(x+y\right)^2}{4\left(z+1\right)}+\frac{\left(y+z\right)^2}{4\left(x+1\right)}+\frac{\left(z+x\right)^2}{4\left(y+1\right)}\)

Đặt \(x+y=p;y+z=q;z+x=r\Rightarrow p+q+r=2\)

\(A\le\Sigma\frac{\left(x+y\right)^2}{4\left(z+1\right)}=\Sigma\frac{\left(x+y\right)^2}{4\left[\left(z+y\right)+\left(z+x\right)\right]}=\frac{p^2}{4\left(q+r\right)}+\frac{r^2}{4\left(p+q\right)}+\frac{q^2}{4\left(p+r\right)}\)

Sau khi đổi biến,cô si thì em ra thế này.Ai đó giúp em với :)

\(\frac{ab}{c+1}=\frac{ab}{a+c+b+c}\le\frac{ab}{4}\left(\frac{1}{a+c}+\frac{1}{b+c}\right)=\frac{1}{4}\left(\frac{ab}{a+c}+\frac{ab}{b+c}\right)\)

\(M\le\frac{1}{4}\left[\frac{ab}{a+c}+\frac{ab}{b+c}+\frac{bc}{a+b}+\frac{bc}{c+a}+\frac{ca}{a+b}+\frac{ca}{b+c}\right]\)

\(=\frac{1}{4}\left[\frac{b\left(a+c\right)}{a+c}+\frac{a\left(b+c\right)}{b+c}+\frac{c\left(a+b\right)}{a+b}\right]=\frac{1}{4}\left(a+b+c\right)=\frac{1}{4}\)

 chịu thôi chị ơi!

Ai trả lời câu này được bái luôn thành sư phụ!!!!!

cảm ơn nha, giờ chị giải được rồi 

Xem lại lời giải đi @Thắng Nguyễn lưu ý ab không phải a^2b^2 :v

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

\(=\sqrt{\frac{ab}{\left(c+a\right)\left(c+b\right)}}\le\frac{1}{2}\left(\frac{ab}{a+c}+\frac{ab}{b+c}\right)\)

Tương tự cho 2 BĐT còn lại r` cộng vào nhé

mấy cái phân số nằm dưới căn thì không âm rồi còn gì

Cho \(a+b+c=1\) nhé các bạn.

Đặt ab + bc + ca = q; abc = r. Ta có:

\(A=\dfrac{\left(ab+bc+ca\right)+6\left(a+b+c\right)+27}{abc+3\left(ab+bc+ca\right)+9\left(a+b+c\right)+27}-\dfrac{1}{3\left(ab+bc+ca\right)}\)

\(A=\dfrac{q+33}{r+3q+36}-\dfrac{1}{3q}\).

Theo bất đẳng thức Schur: \(a^3+b^3+c^3+3abc\ge a^2b+b^2c+c^2a+ab^2+bc^2+ca^2\)

\(\Leftrightarrow\left(a+b+c\right)^3+9abc\ge4\left(a+b+c\right)\left(ab+bc+ca\right)\)

\(\Leftrightarrow9r\ge4q-1\Leftrightarrow r\ge\dfrac{4q-1}{9}\).

Từ đó \(A\le\dfrac{q+33}{\dfrac{4q-1}{9}+3q+36}-\dfrac{1}{3q}\)

\(\Rightarrow A\leq \frac{27q^2+860q-323}{93q^2+969q}\)

\(\Rightarrow A+\dfrac{1}{10}=\dfrac{\left(3q-1\right)\left(121q+3230\right)}{30q\left(31q+323\right)}\le0\). (Do \(q=ab+bc+ca\le\dfrac{\left(a+b+c\right)^2}{3}=\dfrac{1}{3}\))

\(\Rightarrow A\leq \frac{-1}{10}\). Dấu "=" xảy ra khi và chỉ khi a = b = c = 1.

ĐK : a;b;c > 0 

Ta có : \(ab+bc+ac=1\) \(\Leftrightarrow c\left(a+b\right)=1-ab\Leftrightarrow c=\dfrac{1-ab}{a+b}\)

Khi đó :  \(c^2+1=\left(\dfrac{1-ab}{a+b}\right)^2+1\)  \(=\dfrac{\left(ab\right)^2+1+a^2+b^2}{\left(a+b\right)^2}=\dfrac{\left(a^2+1\right)\left(b^2+1\right)}{\left(a+b\right)^2}\)

\(\Rightarrow\dfrac{1}{c^2+1}=\dfrac{\left(a+b\right)^2}{\left(a^2+1\right)\left(b^2+1\right)}\) 

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

Suy ra : \(A=\dfrac{a}{a^2+1}+\dfrac{b}{b^2+1}-\dfrac{1}{c^2+1}=\dfrac{\left(a+b\right)\left(ab+1-a-b\right)}{\left(a^2+1\right)\left(b^2+1\right)}=\dfrac{\left(a+b\right)\left(1-a\right)\left(1-b\right)}{\left(a^2+1\right)\left(b^2+1\right)}\)

AD BĐT Cauchy ta được :  \(\left(a+b\right)\left[\left(1-a\right)\left(1-b\right)\right]\le\dfrac{\left[a+b+\left(1-a\right)\left(1-b\right)\right]^2}{4}=\dfrac{\left(1+ab\right)^2}{4}\)

\(\left(a^2+1\right)\left(b^2+1\right)\ge\left(ab+1\right)^2\)  ( theo BCS )

Suy ra : \(A\le\dfrac{1}{4}\)