1. ## evaluate

1) let $\tan\alpha$ and $\tan\beta$ roots of equation $x^2 + \pi x + \sqrt{2} = 0$ .

Evaluate : $A = \sin^2(\alpha+\beta) + \pi\sin(\alpha+\beta)\cos(\alpha+\beta) + \sqrt{2} \cos^2(\alpha+\beta)$

2) let $x,y,z > 0$, prove that : $\frac{xy}{z} + \frac{yz}{x} + \frac{zx}{y} \geq x + y + z$

3) Find all function $f: \mathbb{R} \to \mathbb{R}$ such that : $f\left( xf(x) + f(y) \right ) = \left(f(x)\right)^2 +y$

2. Originally Posted by linda2005

1) let $\tan\alpha$ and $\tan\beta$ roots of equation $x^2 + \pi x + \sqrt{2} = 0$
[snip]

You can use the quadratic formula to find a and b (alpha and beta).
x=(-B+-sqrt(B^2-4AC))/2A, so x=(-pi+-sqrt(pi^2-4sqrt(2)))/2=tan(angle).
The two possible values of arctan(x) are -.4986 and -1.2032 in radians.
Simply plug these values in for a and b in the second equation.

3. Originally Posted by Media_Man
(1)
You can use the quadratic formula to find a and b (alpha and beta).
x=(-B+-sqrt(B^2-4AC))/2A, so x=(-pi+-sqrt(pi^2-4sqrt(2)))/2=tan(angle).
The two possible values of arctan(x) are -.4986 and -1.2032 in radians.
Simply plug these values in for a and b in the second equation.
thanks media_Man , but we cant use calculator. and we must find exact value of A .

4. ## No problem...

but we cant use calculator. and we must find exact value of A
No problem. Let a=(-pi+sqrt(pi^2-4sqrt(2)))/2 and b=(-pi-sqrt(pi^2-4sqrt(2)))/2. Then a+b=-pi. Verify this for yourself, and I think you can take it from there...

5. Originally Posted by linda2005
[snip]
2) let $x,y,z > 0$, prove that : $\frac{xy}{z} + \frac{yz}{x} + \frac{zx}{y} \geq x + y + z$
[snip]
Pick three random positive numbers x,y,z.

Define vector u=(xy,yz,xz) and vector v=(xz,xy,yz) in R^3

Note: |u|=|v|, so let d=|u|=|v|

Now we ask, which is bigger, u*v or u*u? (where * signifies vector product where appropriate)

Since, u*v=|u||v|cos(a), for a the angle between them,
u*v=d^2cos(a) and u*u=d^2cos(0)=d^2.

Since all values are positive, u and v are confined to the first (+,+,+) quadrant, the maximum angle between them is 90*=pi/2 radians.

Since cosine is a decreasing function on the interval [0,pi/2], cos(a)<cos(0), therefore u*u>u*v. You can also think of the fact that for two vectors of the same length, the dot product increases the closer they are together (the smaller the angle between them), so the maximal dot product for any vector is the dot product with itself.

Substituting, this gives us: (xy)^2+(yz)^2+(xz)^2 > yzx^2+xzy^2+xyz^2. Divide by xyz on both sides and you have your exact answer.

I KNOW THIS FEELS LIKE USING A CANON TO KILL A MOSQUITO, BUT THIS THE LEAST CONFUSING WAY I WAS ABLE TO VISUALIZE THE PROBLEM. I HOPE YOU ARE ABLE TO USE VECTOR GEOMETRY IN YOUR PROOF.

6. Hello, Linda!

1) Let $\tan A$ and $\tan B$ be roots of equation: $x^2 + \pi x + \sqrt{2} \:=\: 0$

Evaluate: . $X \;=\; \sin^2(A+B) + \pi\sin(A+B)\cos(A+B) + \sqrt{2}\cos^2(A+B)$
Since $\tan A,\:\tan B$ are roots of: . $x^2 + \pi x + \sqrt{2} \:=\:0$

. . we have: . $\begin{array}{cccc}\tan A + \tan B &=& -\pi \\ \tan A\tan B &=& \sqrt{2} \end{array}$

Then: . $\tan(A + B) \:=\:\frac{\tan A + \tan B}{1 - \tan A\tan B} \:=\:\frac{-\pi}{1-\sqrt{2}} \quad\Rightarrow\quad \tan(A+B) \:=\:\pi(\sqrt{2}+1)\;\;{\color{blue}[1]}$

We have: . $X \;=\;\sin^2(A+B) + \pi\sin(A+B)\cos(A+B) + \sqrt{2}\cos^2(A+B)$

Divide by $\cos^2(A+B)\!:\;\;\frac{X}{\cos^2(A+B)} \;=\;\frac{\sin^2(A+B)}{\cos^2(A+B)} + \frac{\pi\sin(A+B)\cos(A+B)}{\cos^2(A+B)} \;+$ $\frac{\sqrt{2}\cos^2(A+B)}{\cos^2(A+B)}$

. . . . $X\sec^2(A+B) \;=\;\tan^2(A+B) + \pi\tan(A+B) + \sqrt{2}$

Substitute [1]:
. . $X\sec^2(A+B) \;=\;\bigg[\pi\left(\sqrt{2}+1\right)\bigg]^2 + \pi\bigg[\pi\left(\sqrt{2}+1\right)\bigg] + \sqrt{2} \;=\;\pi^2\left(4+3\sqrt{2}\right) + \sqrt{2}$

. . . . . . . . . $X \;=\;\frac{\pi^2\left(4+3\sqrt{2}\right) + \sqrt{2}}{\sec^2(A+B)}$

Since $\sec^2\!\theta \:=\:\tan^2\!\theta + 1$, we have:
. . $\sec^2(A+B) \;=\;\tan^2(A+B) + 1 \;=\;\bigg[\pi\left(\sqrt{2}+1\right)\bigg]^2 + 1 \;=\;\pi^2\left(3 + 2\sqrt{2}\right)+1$

Therefore: . $X \;=\;\frac{\pi^2\left(4+3\sqrt{2}\right) + \sqrt{2}}{\pi^2\left(3+2\sqrt{2}\right) + 1}$

7. ## Incomplete

Ah, yes. I see what I did wrong.

Your proof is incomplete. Factor out a sqrt(2) from the numerator to give you:

N=sqrt(2)(pi^2(2sqrt(2)+3)+1), which of course cancels with the denominator, leaving X=sqrt(2).

8. Factor out a sqrt(2) from the numerator . . .

Wow! . . . How did I miss that? . . . *blush*

9. Originally Posted by linda2005

2) let $x,y,z > 0$, prove that : $\frac{xy}{z} + \frac{yz}{x} + \frac{zx}{y} \geq x + y + z$
expanding $(a-b)^2 + (b-c)^2 + (c - a)^2 \geq 0$ gives us $a^2 + b^2 + c^2 \geq ab + bc + ac.$ now put $a=xy, \ b = yz, \ c=xz$ and then divide by $xyz.$

3) Find all function $f: \mathbb{R} \to \mathbb{R}$ such that : $f\left( xf(x) + f(y) \right ) = \left(f(x)\right)^2 +y. \ \ \ \ \ (*)$
this question is not bad! if in $(*)$ we put x = 0 we'll get $f(f(y))=(f(0))^2 + y. \ \ \ \ \ \ (**)$

1) $f$ is onto: suppose $z \in \mathbb{R}$ is given. put $y=z - (f(0))^2.$ then by $(**): \ f(f(y))=z.$

2) $f(0)=0$: by 1) there exists $x$ with $f(x)=0.$ then by $(*): \ f(f(y))=y,$ for all $y \in \mathbb{R}.$ let's call this result $(***).$ now $(**)$ gives us $f(0)=0.$

3) $\forall x \in \mathbb{R}: \ (f(x))^2 = x^2$: if in $(*)$ we put y = 0, we'll get $f(xf(x))=(f(x))^2.$ again if in $(*)$ we let y = 0 and replace x with f(x) and use $(***)$ we'll get: $f(xf(x))=x^2.$

4) $f(x)=\pm x$: by 3), for any $x$ either $f(x)=x$ or $f(x)=-x.$ now suppose $f(x)=x$ and $f(y)=-y,$ for some non-zero $x,y.$ then by $(*): \ f(x^2 - y) = x^2 + y.$ on the other hand:

$f(x^2 - y)=\pm(x^2 - y),$ which gives us $x=0$ or $y=0.$ contradiction! therefore either $\forall x \in \mathbb{R}: \ f(x)=x$ or $\forall x \in \mathbb{R}: \ f(x)=-x.$