\(P=a.x^m+b.\frac{1}{x^n}\)

Áp dụng BĐT Co-si cho 2 số dương \(a.x^m\)và \(b.\frac{1}{x^n}\), ta có :

\(a.x^m+b.\frac{1}{x^n}\ge2\sqrt{\frac{ab.x^m}{x^n}}\)

\(\Rightarrow a.x^m+b.\frac{1}{x^n}\ge2\sqrt{ab.x^{m-n}}\)

Vì \(2\sqrt{ab.x^{m-n}}\)Luôn \(\ge0\)\(\Rightarrow\)\(P_{min}=0\Leftrightarrow2\sqrt{ab.x^{m-n}}=0\)

Mà \(a,b>0\Rightarrow x^{m-n}=0\Leftrightarrow m-n=0\Rightarrow m=n\)

Vậy \(P_{min}=0\Leftrightarrow m=n\)

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

Ta có: \(x+y+z=by+cz+ax+cz+ax+by=2\left(ax+by+cz\right)\)Thay \(z=ax+by\)

\(\Rightarrow x+y+z=2\left(z+cz\right)=2z\left(1+c\right)\)

\(\Rightarrow\dfrac{1}{1+c}=\dfrac{2z}{x+y+z}\)

Tương tự:\(\left\{{}\begin{matrix}\dfrac{1}{1+a}=\dfrac{2x}{x+y+z}\\\dfrac{1}{1+b}=\dfrac{2y}{x+y+z}\end{matrix}\right.\)

\(\Rightarrow A=\dfrac{1}{1+a}+\dfrac{1}{1+b}+\dfrac{1}{1+c}=\dfrac{2\left(x+y+z\right)}{x+y+z}=2\)Vậy A=2

Ta có:\(\left\{{}\begin{matrix}x=by+cz\\y=ax+cz\\z=ax+by\end{matrix}\right.\)

\(\Leftrightarrow x+y+z=2\left(ax+by+cz\right)\)

Thay \(x=by+cz\) vào biểu thức ta được:

\(x+y+z=2\left(ax+x\right)=2x\left(a+1\right)\)

\(\Leftrightarrow\dfrac{1}{1+a}=\dfrac{2x}{2x\left(1+a\right)}=\dfrac{2x}{x+y+z}\)

CMTT và cộng theo vế suy ra A=2

Bài 1:
Vì $x+y+z=1$ nên:

\(Q=\frac{x}{x+\sqrt{x(x+y+z)+yz}}+\frac{y}{y+\sqrt{y(x+y+z)+xz}}+\frac{z}{z+\sqrt{z(x+y+z)+xy}}\)

\(Q=\frac{x}{x+\sqrt{(x+y)(x+z)}}+\frac{y}{y+\sqrt{(y+z)(y+x)}}+\frac{z}{z+\sqrt{(z+x)(z+y)}}\)

Áp dụng BĐT Bunhiacopxky:

\(\sqrt{(x+y)(x+z)}=\sqrt{(x+y)(z+x)}\geq \sqrt{(\sqrt{xz}+\sqrt{xy})^2}=\sqrt{xz}+\sqrt{xy}\)

\(\Rightarrow \frac{x}{x+\sqrt{(x+y)(x+z)}}\leq \frac{x}{x+\sqrt{xy}+\sqrt{xz}}=\frac{\sqrt{x}}{\sqrt{x}+\sqrt{y}+\sqrt{z}}\)

Hoàn toàn tương tự với các phân thức còn lại và cộng theo vế suy ra:

\(Q\leq \frac{\sqrt{x}}{\sqrt{x}+\sqrt{y}+\sqrt{z}}+ \frac{\sqrt{y}}{\sqrt{x}+\sqrt{y}+\sqrt{z}}+ \frac{\sqrt{z}}{\sqrt{x}+\sqrt{y}+\sqrt{z}}=1\)

Vậy $Q$ max bằng $1$

Dấu bằng xảy ra khi $x=y=z=\frac{1}{3}$

Bài 2:
Vì $x+y+z=1$ nên:

\(\text{VT}=\frac{1-x^2}{x(x+y+z)+yz}+\frac{1-y^2}{y(x+y+z)+xz}+\frac{1-z^2}{z(x+y+z)+xy}\)

\(\text{VT}=\frac{(x+y+z)^2-x^2}{(x+y)(x+z)}+\frac{(x+y+z)^2-y^2}{(y+z)(y+x)}+\frac{(x+y+z)^2-z^2}{(z+x)(z+y)}\)

\(\text{VT}=\frac{(y+z)[(x+y)+(x+z)]}{(x+y)(x+z)}+\frac{(x+z)[(y+z)+(y+x)]}{(y+z)(y+x)}+\frac{(x+y)[(z+x)+(z+y)]}{(z+x)(z+y)}\)

Áp dụng BĐT AM-GM:
\(\text{VT}\geq \frac{2(y+z)\sqrt{(x+y)(x+z)}}{(x+y)(x+z)}+\frac{2(x+z)\sqrt{(y+z)(y+x)}}{(y+z)(y+x)}+\frac{2(x+y)\sqrt{(z+x)(z+y)}}{(z+x)(z+y)}\)

\(\Leftrightarrow \text{VT}\geq 2\underbrace{\left(\frac{y+z}{\sqrt{(x+y)(x+z)}}+\frac{x+z}{\sqrt{(y+z)(y+x)}}+\frac{x+y}{\sqrt{(z+x)(z+y)}}\right)}_{M}\)

Tiếp tục AM-GM cho 3 số trong ngoặc lớn, suy ra \(M\geq 3\)

Do đó: \(\text{VT}\geq 2.3=6\) (đpcm)

Dấu bằng xảy ra khi $3x=3y=3z=1$

Bài 4:

Ta có một đẳng thức quen thuộc là:

\(1=(a+b)(b+c)(c+a)=(ab+bc+ac)(a+b+c)-abc(*)\)

Mà theo AM-GM:

\((a+b+c)(ab+bc+ac)\geq 3\sqrt[3]{abc}.3\sqrt[3]{ab.bc.ac}=9abc\)

\(\Rightarrow abc\leq \frac{(a+b+c)(ab+bc+ac)}{9}(**)\)

Từ \((*);(**)\Rightarrow 1\geq \frac{8}{9}(a+b+c)(ab+bc+ac)\)

Theo tính chất quen thuộc của BĐT AM-GM:

\((a+b+c)^2\geq 3(ab+bc+ac)\Rightarrow a+b+c\geq \sqrt{3(ab+bc+ac)}\)

Do đó:

\(1\geq \frac{8}{9}\sqrt{3(ab+bc+ac)^3}\)

\(\Rightarrow (ab+bc+ac)^3\leq \frac{27}{64}\Rightarrow ab+bc+ac\leq \frac{3}{4}\)

Ta có đpcm