In this note, I have a selection of questions for you to prove. It is for all levels to try. The theme is algebra. Good luck.

1) Prove that

$\dfrac {a + b}{c^2} + \dfrac {a + c}{b^2} + \dfrac {b + c}{a^2} \geq 2 \left ( \dfrac {1}{a} + \dfrac {1}{b} + \dfrac {1}{c} \right )$

for any positive real numbers $a$, $b$ and $c$.

2) Find all functions $f : \mathbb{R} \rightarrow \mathbb{R}$ such that

$f(x^2 - y^2) = x^2 - (f(y))^2$

for all real numbers $x$ and $y$.

3) Consider the polynomial $p(x) = x^3 - 3x + 1$

(a) Prove that the equation $p(x) = 0$ has exactly three real solutions.

(b) If the solutions to $p(x) = 0$ are $\alpha$, $\beta$and $\gamma$, find a monic cubic polynomial whose roots are $\dfrac {1}{\alpha^2}$, $\dfrac {1}{\beta^2}$ and $\dfrac {1}{\gamma^2}$.

4) The equation $x^4 + ax^3 + bx^2 + cx + d = 0$ has four real positive roots.

Prove that

(a) $ac \geq 16d$

(b) $b^2 \geq 36d$

5) A sequence of real functions $f_n$ is defined by

$f_1(x) = \sqrt{x^2 + 48}$

and

$f_{n + 1}(x) = \sqrt{x^2 + 6 f_n(x)}$

for $n \geq 1$.

Find all real solutions of the equation $f_{2007}(x) = 2x$

6) Let $n$ be a positive integer. Suppose that $a_1, a_2, \ldots, a_n, b_1, b_2, \ldots, b_n$ are positive real numbers such that

$\displaystyle \sum_{i = 1}^n a_i = \displaystyle \sum_{i = 1}^n b_n$

Prove that

$\displaystyle \sum_{i = 1}^n \dfrac {a_i^2}{a_i + b_i} \geq \dfrac {1}{2} \displaystyle \sum_{i = 1}^n a_i$

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## Comments

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TopNewestLet $P(x,y)$ denote the equation. Then $P(0,0)\implies f(0)\in\{0,-1\}$. If $f(0)=0$ then for all non negative $x$ we have $P(\sqrt x,0)\implies f(x)=x$. Now $P(0,\sqrt y)\implies f(-y)=-y$ for all non negative $y$. Thus $f(x)=x$ is a solution for all $x$, which clearly works. If $f(0)=-1$ then $P(\sqrt x,0)\implies x-1$ for all non negative $x$. But choosing positive reals $x>y$ we see that this doesn't satisfy $P(x,y)$. So only solution is $f(x)=x~\forall x\in\mathbb R$.

The discriminant of the given cubic is $81>0$, so the roots are all real and distinct. Let our desired cubic be $q(x)=x^3-px^2+qx-r$. By Vieta's we must have $p=\alpha^2\beta^2+\beta^2\gamma^2+\gamma^2\alpha^2$, $q=\alpha^2+\beta^2+\gamma^2$ and $r=1$. Using the Vieta equations $\alpha+\beta+\gamma=0$ and $\alpha\beta+\beta\gamma+\gamma\alpha=-3$ we can easily find $\alpha^2+\beta^2+\gamma^2=6$ and $\alpha^2\beta^2+\beta^2\gamma^2+\gamma^2\alpha^2=9$. So $q(x)=x^3-9x^2+6x-1$.

If the roots are $p,q,r,s$, by Vieta's $ac\ge 16d~\Leftrightarrow ~(p+q+r+s)(pqr+pqs+prs+qrs)\ge 16pqrs$ which follows by applying AM-GM on each bracket. Also $b^2\ge 36d ~\Leftrightarrow (pq+pr+ps+qr+qs+rs)^2\ge 36pqrs$, again follows directly by applying the AM-GM inequality inside.

Notice that $f_1(x)=2x$ has only solution $x=4$ which, by induction, is the root of all $f_n(x)-2x$. It easily follows that $f_1(x)-2x$ is monotonically decreasing. So by induction all of $f_n(x)-2x$, being compositions of monotone mappings, are monotonically decreasing. So each $f_n(x)-2x$ has only one root, which is $4$.

Direct application of Titu's Lemma. We have $\large\sum \dfrac{a_i^2}{a_i+b_i}\ge \dfrac{\left(\sum a_i\right)^2}{\sum a_i+\sum b_i}=\dfrac{\left(\sum a_i\right)^2}{2\sum a_i}=\dfrac{1}{2}\sum a_i.$

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How you calculated discriminant of a cubic please help me

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Go to the link attached to the word discriminant in the solution.

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The last one is quite easy.

$\displaystyle \bigg( \sum_{i=1}^n \frac{a_i ^2}{a_i + b_i} \bigg) \bigg( \sum_{i=1}^n (a_i + b_i) \bigg) \geq \bigg( \sum_{i=1}^n a_i \bigg)^2$

$\displaystyle \bigg( \sum_{i=1}^n \frac{a_i ^2}{a_i + b_i} \bigg) \geq \bigg( \frac{(\sum_{i=1}^n a_i)^2}{ 2 \sum_{i=1}^n a_i } \bigg) \Rightarrow \sum_{i=1}^n \frac{a_i ^2}{a_i + b_i} \geq \frac{1}{2} \sum_{i=1}^n a_i$

It is a straightforward application of Cauchy-Schwarz Inequality.

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Here is another solution to the last one. That was how I approached it a long time ago, and this method stuck with me.

Observe that $\sum \frac{ a_i ^2 - b_i ^ 2 } { a_i + b_i} = \sum (a_i - b_i) = 0$.

Hence, this gives us $\sum \frac{ a_i ^ 2 } {a_i + b_i } = \sum \frac { b_i^2 } { a_i + b_i }$.

We want to show that $\sum \frac{ a_i^2 + b_i^2 } { a_i + b_i} = 2 \sum \frac { a_i^2 } { a_i + b_i } \geq \sum a_i$.

By QM-AM, we know that $\sqrt{ \frac{ a_i ^2 + b_i^2 } { 2} } \geq \frac{ a_i + b_i} { 2}$, or that $\frac{ a_i^2 + b_i^2} {a_i + b_i } \geq \frac{ a_i + b_i} {2}$.

Hence, we get that $\sum \frac{ a_i ^2 + b_i^2 } { a_i + b_i } \geq \sum \frac{ a_i + b_i } { 2} = \sum a_i$, and we are done.

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Solution to the first problem -

adding $\frac{1}{a} + \frac{1}{b} + \frac{1}{c}$ to both sides , we get -

$\frac {(a + b+ c)}{3}(\frac{1}{a^{2}} + \frac{1}{b^{2}} + \frac{1}{c^{2}}) > (\frac{1}{a} + \frac{1}{b} + \frac{1}{c})$

since $\frac{a + b + c}{3} > \frac{3}{(\frac{1}{a} + \frac{1}{b} +\frac{1}{c})}$

the problem reduces to proving -

$3(\frac{1}{a^{2}} + \frac{1}{b^{2}} + \frac{1}{c^{2}}) > (\frac{1}{a} + \frac{1}{b} + \frac{1}{c})^{2}$

which directly follows from Cauchy - Schwarz inequality.

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For 2nd problem I guess there are many

$\displaystyle f(x^{2} - y^{2}) = x^{2} + (f(y))^{2}$

Put x = 0

$\displaystyle f(-y^{2}) = (f(y))^{2}$

Now here I have a doubt that it is correct or not

$\displaystyle f(y) = -y^{n}, n \in \mathbb {ODD}$

There can be $\infty$ functions for different 'n'

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Try putting the function you found in the first condition. You only solved for the case when $x = 0$.

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Yes but in that case only I am getting $\infty$ functions.

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$f(x) = - x^2$ does not satisfy the original functional equation.

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N should be odd

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$f(x) = - x^3$ does not satisfy the original functional equation.

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I think you did something wrong. It asks for $f(x^2 - y^2) = x^2 - (f(y))^2$.

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Good

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What do you mean by that? Can you post your solutions, if you have any.

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