Question

Chứng minh rằng

a)a²+b²+c²+d²+m²-a(b+c+d+m)\(\ge\)0 với mọi a,b,c,d,m

b)\(\dfrac{1}{x}+\dfrac{1}{y}\ge\dfrac{4}{x+y}\)(x;y>0)

c)(ab+cd)²\(\le\)(a²+c²)(b²+d²)

d)a²+b²\(\ge\)a+b-\(\dfrac{1}{2}\)

Answer 1

a)

\(a^2+b^2+c^2+d^2+m^2-a(b+c+d+m)\)

\(=\frac{4a^2+4b^2+4c^2+4d^2+4m^2-4a(b+c+d+m)}{4}\)

\(=\frac{(a^2+4b^2-4ab)+(a^2+4c^2-4ac)+(a^2+4d^2-4ad)+(a^2+4m^2-4am)}{4}\)

\(=\frac{(a-2b)^2+(a-2c)^2+(a-2d)^2+(a-2m)^2}{4}\geq 0\) (đpcm)

Dấu "=" xảy ra khi \(a=2b=2c=2d=2m\)

b)

Xét hiệu

\(\frac{1}{x}+\frac{1}{y}-\frac{4}{x+y}=\frac{x+y}{xy}-\frac{4}{x+y}=\frac{(x+y)^2-4xy}{xy(x+y)}\)

\(=\frac{x^2+y^2-2xy}{xy(x+y)}=\frac{(x-y)^2}{xy(x+y)}\geq 0, \forall x,y>0\)

\(\Rightarrow \frac{1}{x}+\frac{1}{y}\geq \frac{4}{x+y}\) (đpcm)

Dấu "=" xảy ra khi $x=y$

Answer 2

c)

Xét hiệu:

\((a^2+c^2)(b^2+d^2)-(ab+cd)^2\)

\(=(a^2b^2+a^2d^2+c^2b^2+c^2d^2)-(a^2b^2+2abcd+c^2d^2)\)

\(=a^2d^2-2abcd+b^2c^2=(ad-bc)^2\geq 0\)

\(\Rightarrow (a^2+c^2)(b^2+d^2)\geq (ab+cd)^2\) (đpcm)

Dấu "=" xảy ra khi \(ad=bc\)

d)

Xét hiệu:

\(a^2+b^2-(a+b-\frac{1}{2})=a^2+b^2-a-b+\frac{1}{2}\)

\(=(a^2-a+\frac{1}{4})+(b^2-b+\frac{1}{4})\)

\(=(a-\frac{1}{2})^2+(b-\frac{1}{2})^2\geq 0\)

\(\Rightarrow a^2+b^2\geq a+b-\frac{1}{2}\) (đpcm)

Dấu "=" xảy ra khi \(a=b=\frac{1}{2}\)