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# Algebraic Proofs

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

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

Note by Sharky Kesa
2 years, 11 months ago

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1. Clearing denominators leads to $$[3,2,0]\ge [2,2,1]$$, true by Muirhead's since $$\{3,2,0\}\succ \{2,2,1\}$$.

2. Let $$P(x,y)$$ 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 $$f(x)=x~\forall x\in\mathbb R$$.

3. 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 $$q(x)=x^3-9x^2+6x-1$$.

4. 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.

5. 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 $$4$$.

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

- 2 years, 11 months ago

- 2 years, 11 months ago

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

- 2 years, 11 months ago

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.

- 2 years, 11 months ago

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.

Staff - 2 years, 11 months ago

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.

- 2 years, 10 months ago

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'

- 2 years, 11 months ago

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

- 2 years, 11 months ago

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

- 2 years, 11 months ago

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

- 2 years, 11 months ago

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

Staff - 2 years, 11 months ago

N should be odd

- 2 years, 11 months ago

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

Staff - 2 years, 11 months ago

Ok so the problem is making a case x= 0?

- 2 years, 11 months ago

Good

- 2 years, 11 months ago

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

- 2 years, 11 months ago