Xét \(\dfrac{a}{a^2+1}+\dfrac{3\left(a-2\right)}{25}-\dfrac{2}{5}=\dfrac{a}{a^2+1}+\dfrac{3a-16}{25}=\dfrac{\left(3a-4\right)\left(a-2\right)^2}{25\left(a^2+1\right)}\ge0\)

\(\Rightarrow\dfrac{a}{a^2+1}\ge\dfrac{2}{5}-\dfrac{3\left(a-2\right)}{25}\)

CMTT \(\Rightarrow\left\{{}\begin{matrix}\dfrac{b}{b^2+1}\ge\dfrac{2}{5}-\dfrac{3\left(b-2\right)}{25}\\\dfrac{c}{c^2+1}\ge\dfrac{2}{5}-\dfrac{3\left(c-2\right)}{25}\end{matrix}\right.\)

Cộng vế theo vế:

\(\Rightarrow VT\ge\dfrac{2}{5}+\dfrac{2}{5}+\dfrac{2}{5}-\dfrac{3\left(a-2\right)+3\left(b-2\right)+3\left(c-2\right)}{25}\ge\dfrac{6}{5}-\dfrac{3\left(a+b+c-6\right)}{25}=\dfrac{6}{5}\)

Dấu \("="\Leftrightarrow a=b=c=2\)

 Bạn tham khảo bài này trên Quanda nha.  

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

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

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

Dấu = xảy ra khi a=b=c=1

1. Vì x, y, z > 0

\(xy+yz+zx\ge2xyz\)

\(\Leftrightarrow\dfrac{1}{x}+\dfrac{1}{y}+\dfrac{1}{z}\ge2\)

Suy ra:

\(\dfrac{1}{x}\ge1-\dfrac{1}{y}+1-\dfrac{1}{z}=\dfrac{y-1}{y}+\dfrac{z-1}{z}\ge2\sqrt{\dfrac{\left(y-1\right)\left(z-1\right)}{yz}}\). (1)

Tương tự \(\dfrac{1}{y}\ge2\sqrt{\dfrac{\left(z-1\right)\left(x-1\right)}{zx}}\) (2)

và \(\dfrac{1}{z}\ge2\sqrt{\dfrac{\left(x-1\right)\left(y-1\right)}{xy}}\) (3)

Nhân (1), (2), (3) với nhau theo vế ta được

\(\dfrac{1}{xyz}\ge\dfrac{8\left(x-1\right)\left(y-1\right)\left(z-1\right)}{xyz}\)

\(\Leftrightarrow\left(x-1\right)\left(y-1\right)\left(z-1\right)\le\dfrac{1}{8}\)

Đẳng thức xảy ra \(\Leftrightarrow x=y=z=\dfrac{3}{2}\)

\(\dfrac{c+1}{c+3}\ge\dfrac{1}{a+2}+\dfrac{3}{b+4}\)

\(\Leftrightarrow1-\dfrac{2}{c+3}\ge\dfrac{1}{a+2}+\dfrac{3}{b+4}\)

\(\Leftrightarrow1-\dfrac{1}{a+2}\ge\dfrac{3}{b+4}+\dfrac{2}{c+3}\ge2\sqrt{\dfrac{6}{\left(b+4\right)\left(c+3\right)}}\)

Hay \(\dfrac{a+1}{a+2}\ge2\sqrt{\dfrac{6}{\left(b+4\right)\left(c+3\right)}}\) (1)

Tương tự \(\dfrac{b+1}{b+4}\ge2\sqrt{\dfrac{2}{\left(c+3\right)\left(a+2\right)}}\) (2)

và \(\dfrac{c+1}{c+3}\ge2\sqrt{\dfrac{3}{\left(a+2\right)\left(b+4\right)}}\) (3)

Nhân (1), (2), (3) vế theo vế

\(\dfrac{\left(a+1\right)\left(b+1\right)\left(c+1\right)}{\left(a+2\right)\left(b+4\right)\left(c+3\right)}\ge8.\dfrac{6}{\left(a+2\right)\left(b+4\right)\left(c+3\right)}\)

\(\Leftrightarrow\left(a+1\right)\left(b+1\right)\left(c+1\right)\ge48\)

Đẳng thức xảy ra \(\Leftrightarrow\left\{{}\begin{matrix}a=1\\b=5\\c=3\end{matrix}\right.\)

\(a+b+c+2=abc\)

\(\Leftrightarrow2a+2b+2c+3+ab+bc+ca=abc+ab+bc+ca+a+b+c+1\)

\(\Leftrightarrow\left(a+1\right)\left(b+1\right)+\left(c+1\right)\left(b+1\right)+\left(c+1\right)\left(a+1\right)=\left(a+1\right)\left(b+1\right)\left(c+1\right)\)

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

Đặt \(\left(\dfrac{1}{a+1};\dfrac{1}{b+1};\dfrac{1}{c+1}\right)=\left(x;y;z\right)\)

\(\Rightarrow x+y+z=1\)

BĐT trở thành:

\(P=\sqrt{x}+\sqrt{y}+\sqrt{z}\le\sqrt{3\left(x+y+z\right)}=\sqrt{3}\)

Dấu "=" xảy ra khi và chỉ khi \(x=y=z=\dfrac{1}{3}\) hay \(a=b=c=2\)

mong mn giúp mk vs 

\(1,yz\sqrt{x-1}=yz\sqrt{\left(x-1\right)\cdot1}\le yz\cdot\dfrac{x-1+1}{2}=\dfrac{xyz}{2}\)

\(zx\sqrt{y-2}=\dfrac{zx\cdot2\sqrt{2\left(y-2\right)}}{2\sqrt{2}}\le\dfrac{xyz}{2\sqrt{2}}\\ xy\sqrt{z-3}=\dfrac{xy\cdot2\sqrt{3\left(z-3\right)}}{2\sqrt{3}}\le\dfrac{xyz}{2\sqrt{3}}\)

\(\Leftrightarrow M\le\dfrac{\dfrac{xyz}{2}+\dfrac{xyz}{2\sqrt{2}}+\dfrac{xyz}{2\sqrt{3}}}{xyz}=\dfrac{xyz\left(\dfrac{1}{2}+\dfrac{1}{2\sqrt{2}}+\dfrac{1}{2\sqrt{3}}\right)}{xyz}=\dfrac{1}{2}+\dfrac{1}{2\sqrt{2}}+\dfrac{1}{2\sqrt{3}}\)

Dấu \("="\Leftrightarrow\left\{{}\begin{matrix}x-1=1\\y-2=2\\z-3=3\end{matrix}\right.\Leftrightarrow\left\{{}\begin{matrix}x=2\\y=4\\z=6\end{matrix}\right.\)

\(2,N^2=\left(\sqrt{a+b}+\sqrt{b+c}+\sqrt{c+a}\right)^2\\ \Leftrightarrow N^2\le\left(a+b+b+c+c+a\right)\left(1^2+1^2+1^2\right)\\ \Leftrightarrow N^2\le6\left(a+b+c\right)=6\sqrt{2}\\ \Leftrightarrow N\le\sqrt{6\sqrt{2}}\)

Dấu \("="\Leftrightarrow a=b=c=\dfrac{\sqrt{2}}{3}\)