Question

a) Với \(n\in N\). Chứng minh:

\(\sqrt{\left(n+1\right)^2}+\sqrt{n^2}=\left(n+1\right)^2-n^2\)

b) Cho a,b,c > 0. Chứng minh:

+) Nếu \(a+b+c=\sqrt{ab}+\sqrt{bc}+\sqrt{ca}\) thì a = b = c.

+) \(\dfrac{a}{b}+\dfrac{b}{c}+\dfrac{c}{a}\ge\sqrt{\dfrac{a}{c}}+\sqrt{\dfrac{b}{a}}+\sqrt{\dfrac{c}{b}}\).

Answer 1

a) CM:\(\sqrt{\left(n+1\right)^2}+\sqrt{n^2}=\left(n+1\right)^2-n^2\)

\(\Leftrightarrow n+1+n=\left(n+1-n\right)\left(n+1+n\right)\)

\(\Leftrightarrow2n+1=1\left(2n+1\right)\)

\(\Leftrightarrow2n+1=2n+1\)

\(\Rightarrow\sqrt{\left(n+1\right)^2}+\sqrt{n^2}=\left(n+1\right)^2-n^2\)

Answer 2

Câu b) ý 2:

Áp dụng BĐT cô si ta có :

\(\dfrac{a}{b}+\dfrac{b}{c}\ge2\sqrt{\dfrac{a}{c}}\\ \dfrac{b}{c}+\dfrac{c}{a}\ge2\sqrt{\dfrac{b}{a}}\\ \dfrac{c}{a}+\dfrac{a}{b}\ge2\sqrt{\dfrac{c}{b}}\\ \Leftrightarrow2\left(\dfrac{a}{b}+\dfrac{b}{c}+\dfrac{c}{a}\right)\ge2\left(\sqrt{\dfrac{a}{c}}+\sqrt{\dfrac{b}{a}}+\sqrt{\dfrac{c}{b}}\right)\\ \Rightarrowđpcm\)

Answer 3

Câu a:

VT=n+1+n=2n+1 (1)

\(VP=n^2+2n+1-n^2=2n+1\) (2)

Từ (1) và (2) => VT=VP =>đpcm

Answer 4

Áp dụng BĐT Côsi ta có :

\(a+b\ge2\sqrt{ab}\\ b+c\ge2\sqrt{bc}\\ c+a\ge2\sqrt{ca}\\ \Leftrightarrow2\left(a+b+c\right)\ge2\left(\sqrt{ab}+\sqrt{bc}+\sqrt{ca}\right)\\ \Leftrightarrow a+b+c\ge\sqrt{ab}+\sqrt{bc}+\sqrt{ca}\)

Dấu bằng xảy ra <=> a=b=c

Vậy nếu \(a+b+c=\sqrt{ab}+\sqrt{bc}+\sqrt{ca}\) thì a=b=c (đpcm)