1.Cho x, y \(\ge\)0 và x+ y=1

Chứng minh rằng : \(x^3+y^3\ge\dfrac{1}{4}\)

2. Cho \(a,b,c\ge0\).Chứng minh rằng:

a, \(a^3+b^3>ab\left(a+b\right)\)

b, \(a^3+b^3+c^3\ge a^2b+ b^2c+c^2a\)

3. Cho x+ y+ z=3 và x, y, z>0. Chứng minh rằng:

a, \(P=\dfrac{1}{x+1}+\dfrac{1}{y+1}+\dfrac{1}{z+1}\ge\dfrac{3}{2}\)

b, \(Q=\dfrac{x}{x^2+1}+\dfrac{y}{y^2+1}+\dfrac{z}{z^2+1}\le\dfrac{3}{2}\)

Đề bài: ax,y,z >0 và \(\sqrt{x}+\sqrt{y}+\sqrt{z}=1\). Tìm Min P= \(\dfrac{x^3}{y+z}+\dfrac{y^3}{z+x}+\dfrac{z^3}{x+y}\).

ĐÁP ÁN:

Ta có: \(\dfrac{x^3}{y+z}+\dfrac{y+z}{36}+\dfrac{1}{162}+\dfrac{y^3}{x+z}+\dfrac{x+z}{36}+\dfrac{1}{162}+\dfrac{z^3}{x+y}+\dfrac{x+y}{36}+\dfrac{1}{162}\ge3\sqrt[3]{\dfrac{x^3}{y+z}.\dfrac{y+z}{36}.\dfrac{1}{162}}+3\sqrt[3]{\dfrac{y^3}{x+z}.\dfrac{x+z}{36}.\dfrac{1}{162}}+3\sqrt[3]{\dfrac{z^3}{x+y}.\dfrac{x+y}{36}.\dfrac{1}{162}}=3\sqrt[3]{\dfrac{x^3}{36.162}}+3\sqrt[3]{\dfrac{y^3}{36.162}}+3\sqrt[3]{\dfrac{z^3}{36.162}}=\dfrac{x+y+z}{6}.\)

=> P+\(\dfrac{x+y+z}{18}+\dfrac{1}{54}\)≥\(\dfrac{x+y+z}{6}\) <=> P≥\(\dfrac{x+y+z}{6}-\dfrac{x+y+z}{18}-\dfrac{1}{54}\)=\(\dfrac{x+y+z}{9}-\dfrac{1}{54}\)

Ta c/m đc: 3(x+y+z)≥(\(\sqrt{x}+\sqrt{y}+\sqrt{z}\))² <=> 2(x+y+z) ≥2\(\left(\sqrt{xy}+\sqrt{xz}+\sqrt{yz}\right)\)<=> x+y+z≥\(\sqrt{xy}+\sqrt{xz}+\sqrt{yz}\)(luôn đúng)

➩x+y+z ≥ \(\dfrac{\left(\sqrt{x}+\sqrt{y}+\sqrt{z}\right)^3}{3}=\dfrac{1}{3}\) => P≥\(\dfrac{1}{54}\). Dấu ''='' xảy ra <=> x=y=z=\(\dfrac{1}{9}\)

\(\dfrac{x^2}{y+z}+\dfrac{y^2}{z+x}+\dfrac{z^2}{x+y}\)

do x,y,z≥0 nên x²≥0 , y+z≥0

áp dụng bất đẳng thức cosi cho 2 số dương \(\dfrac{x^2}{y+z}\) và y+z/4

x^2/y+z +(y+z)/4≥2\(\sqrt{\dfrac{x^2}{y+z}.\dfrac{\left(y+z\right)}{4}}\) =x (1)

y^2/x+z+(x+z)/4≥2\(\sqrt{\dfrac{y^2}{x+z}.\dfrac{x+z}{4}}\) =y (2)

z^2/y+x+(y+x)/4≥2\(\sqrt{\dfrac{z^2}{y+x}.\dfrac{y+x}{4}}\) =z (3)

từ (1)(2)(3)

➜\(\dfrac{x^2}{y+z}+\dfrac{y^2}{z+x}+\dfrac{z^2}{x+y}\)+(y+z/4)+(z+x)/4+(x+y)/4 ≥ x+y+z

⇔\(\dfrac{x^2}{y+z}+\dfrac{y^2}{z+x}+\dfrac{z^2}{x+y}\) +(a+b+c)/2 ≥x+y+z

⇔\(\dfrac{x^2}{y+z}+\dfrac{y^2}{z+x}+\dfrac{z^2}{x+y}\) ≥ (x+y+z)/2

⇔\(\dfrac{x^2}{y+z}+\dfrac{y^2}{z+x}+\dfrac{z^2}{x+y}\) ≥1 (vì x+y+z=2)

vậy giá trị nhỏ nhất của \(\dfrac{x^2}{y+z}+\dfrac{y^2}{z+x}+\dfrac{z^2}{x+y}\) =1

bài 1: giải các hệ phương trình

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

x+y=9

2) \(\dfrac{2x+1}{4}-\dfrac{y-2}{3}=\dfrac{1}{12}\)

\(\dfrac{x+5}{2}-\dfrac{y+7}{3}=-4\)

3)\(2|x|-y=3\)

\(|x|+y=3\)

4)\(2\left(x+y\right)+\sqrt{x+1}=4\)

\(\left(x+y\right)-3\sqrt{x+1}=-5\)

5) \(\dfrac{7}{2x+y}+\dfrac{4}{2x-y}=74\)

\(\dfrac{3}{2x+y}+\dfrac{2}{2x-y}=32\)

6)\(\dfrac{1}{x}+\dfrac{3}{2y+1}=2\)

\(\dfrac{2}{x}+\dfrac{4}{2y+1}=2\)

7) \(\dfrac{1}{x}+\dfrac{1}{y}=2\)

\(\dfrac{3}{x}-\dfrac{1}{y}=2\)

8)\(\dfrac{1}{x+2}+\dfrac{3}{2y-1}=4\)

\(\dfrac{4}{x+2}-\dfrac{1}{2y-1}=3\)

9)\(\dfrac{4}{x+y} +\dfrac{1}{y-1}=5\)

\(\dfrac{1}{x+y}-\dfrac{2}{y-1}=-1\)

10)\(\dfrac{7}{\sqrt{2x+3}}-\dfrac{4}{\sqrt{3}-y}=\dfrac{5}{3}\)

\(\dfrac{5}{\sqrt{2x+3}}+\dfrac{3}{\sqrt{3-y}}=\dfrac{13}{6}\)

11)\(\dfrac{3x}{x-1}-\dfrac{2}{y+2}=4\)

\(\dfrac{2x}{x-1}+\dfrac{1}{y+2}=5\)

12) \(\dfrac{7}{\sqrt{x}-7}-\dfrac{4}{\sqrt{y}+6}=\dfrac{5}{3}\)

\(\dfrac{5}{\sqrt{x}-7}+\dfrac{3}{\sqrt{y}+6}2\dfrac{1}{6}\)

13) \(3\sqrt{x-1}+2\sqrt{y}=13\)

\(2\sqrt{x-1}-\sqrt{y}=4\)

14) 6x + 6y = 5xy

\(\dfrac{4}{x}-\dfrac{3}{y}=1\)

a) Với 0 < x <\(\dfrac{4}{3}\), chứng minh rằng \(\dfrac{1}{x^2\left(4-3x\right)}\) \(\ge\) x

b) Cho a,b,c là ba số dương nhỏ hơn \(\dfrac{4}{3}\) sao cho a + b + c = 3. Chứng minh rằng:

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

bài 2: rút gọn

a) \(\dfrac{3+\sqrt{3}}{\sqrt{3}}+\dfrac{\sqrt{6}-\sqrt{3}}{1-\sqrt{2}}\)

b) \(\sqrt{\dfrac{x-2\sqrt{x}+1}{x+2\sqrt{x}+1}}\) (x\(\ge\) 0

c) \(\dfrac{x-y}{\sqrt{y}-1}\sqrt{\dfrac{\left(y-2\sqrt{x}+1\right)^2}{\left(x-1\right)^4}}\) x \(\ne\) 1; y \(\ne\)1 ; y \(\ge\)0