Printable View
Hello everyone.
I would like to start 'marathon' of the exercises from different competition in the secondary school level. (ages 15-18)
let's start with inequality :)
Exercise 1.
Let $\displaystyle a,b$ be positive real numbers. Prove that
$\displaystyle \left(1+\frac{a}{b} \right)^m+\left(1+\frac{b}{a} \right)^m \ge 2^{m+1}$
where $\displaystyle m$ is natural.
Enjoy!
If $\displaystyle a=b,$ we have the equality so assume, wlog, that $\displaystyle a>b$ and that $\displaystyle a=kb, \ \ k>1.$
We then have to prove that
$\displaystyle (1+k)^{m}+(1+1/k)^{m} > 2^{m+1}.$
Now, $\displaystyle [(1+k)^{m/2}-(1+1/k)^{m/2}]^{2}=(1+k)^{m}+(1+1/k)^{m}-2(1+k)^{m/2}(1+1/k)^{m/2}>0,$
so,
$\displaystyle (1+k)^{m}+(1+1/k)^{m}>2(1+k)^{m/2}(1+1/k)^{m/2}=2[(1+k)(1+1/k)]^{m/2}$
$\displaystyle =2(2+k+1/k)^{m/2}.$
If $\displaystyle k>2,$ then $\displaystyle 2+k+1/k>4+1/k=2^{2}+1/k$ and the result follows, so assume that $\displaystyle 1<k<2$ and that $\displaystyle k=1+\alpha$ with $\displaystyle 0<\alpha<1.$
Then,
$\displaystyle 2+k+1/k=2+(1+\alpha)+1/(1+\alpha)=3+\alpha+1-\alpha+\alpha^{2}-...$,
$\displaystyle =4+\alpha^{2}-\alpha^{3}+...>2^{2},$ and again the result follows.
You could a bit easier: AM-GM two times. But your solution is also OK. Now is your turn - send your exercise.
O.K.
Exercise 2.
Prove that there do not exist positive integers $\displaystyle x, y, z$ such that
$\displaystyle x^2 +y^2 + z^2 = 2xyz.$
$\displaystyle x^2 +y^2 + z^2 = 2xyz$
$\displaystyle (x+y+z)^2=2(x+1)(y+1)(z+1)-2(x+y+z)-2$
$\displaystyle (x+y+z)^2+2(x+y+z)=2(x+1)(y+1)(z+1)-2$
$\displaystyle (x+y+z)(x+y+z+2)=2[(x+1)(y+1)(z+1)-1]$
what is impossible.
Exercise 3.
Calculate
$\displaystyle \sqrt{ \underbrace{44...4}_{2n}+\underbrace{22...2}_{n+1} +\underbrace{88...8}_{n}+7}$
