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

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

Markdown

Appears as

*italics* or _italics_

italics

**bold** or __bold__

bold


- bulleted
- list

bulleted
list


1. numbered
2. list

numbered
list

Note: you must add a full line of space before and after lists for them to show up correctly

paragraph 1

paragraph 2

paragraph 1

paragraph 2

[example link](https://brilliant.org)

example link

> This is a quote

This is a quote

    # I indented these lines
    # 4 spaces, and now they show
    # up as a code block.

    print "hello world"

# I indented these lines
# 4 spaces, and now they show
# up as a code block.

print "hello world"

Math

Appears as

Remember to wrap math in $ ... $ or \[ ... \] to ensure proper formatting.

2 \times 3

2 \times 3

2^{34}

2^{34}

a_{i-1}

a_{i-1}

\frac{2}{3}

\frac{2}{3}

\sqrt{2}

\sqrt{2}

\sum_{i=1}^3

\sum_{i=1}^3

\sin \theta

\sin \theta

\boxed{123}

\boxed{123}

Easy Math Editor

This discussion board is a place to discuss our Daily Challenges and the math and science related to those challenges. Explanations are more than just a solution — they should explain the steps and thinking strategies that you used to obtain the solution. Comments should further the discussion of math and science.

When posting on Brilliant:

Use the emojis to react to an explanation, whether you're congratulating a job well done , or just really confused .
Ask specific questions about the challenge or the steps in somebody's explanation. Well-posed questions can add a lot to the discussion, but posting "I don't understand!" doesn't help anyone.
Try to contribute something new to the discussion, whether it is an extension, generalization or other idea related to the challenge.
Stay on topic — we're all here to learn more about math and science, not to hear about your favorite get-rich-quick scheme or current world events.

Markdown	Appears as
`italics` or `_italics_`	italics
`bold` or `__bold__`	bold
- bulleted - list	bulleted list
1. numbered 2. list	numbered list
Note: you must add a full line of space before and after lists for them to show up correctly
paragraph 1 paragraph 2	paragraph 1 paragraph 2
`[example link](https://brilliant.org)`	example link
`> This is a quote`	This is a quote
# I indented these lines # 4 spaces, and now they show # up as a code block. print "hello world"	# I indented these lines # 4 spaces, and now they show # up as a code block. print "hello world"

Math	Appears as
Remember to wrap math in `$` ... `$` or `\[` ... `\]` to ensure proper formatting.
`2 \times 3`	$2 \times 3$
`2^{34}`	$2^{34}$
`a_{i-1}`	$a_{i-1}$
`\frac{2}{3}`	$\frac{2}{3}$
`\sqrt{2}`	$\sqrt{2}$
`\sum_{i=1}^3`	$\sum_{i=1}^3$
`\sin \theta`	$\sin \theta$
`\boxed{123}`	$\boxed{123}$

Comments

Clearing denominators leads to $[3,2,0]\ge [2,2,1]$ , true by Muirhead's since $\{3,2,0\}\succ \{2,2,1\}$ .
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 $f(x)=x~\forall x\in\mathbb R$ .
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 $\large\sum \dfrac{a_i^2}{a_i+b_i}\ge \dfrac{\left(\sum a_i\right)^2}{\sum a_i+\sum b_i}=\dfrac{\left(\sum a_i\right)^2}{2\sum a_i}=\dfrac{1}{2}\sum a_i.$

Jubayer Nirjhor - 6 years, 6 months ago

How you calculated discriminant of a cubic please help me

Aman Sharma - 6 years, 6 months ago

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

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.

Sudeep Salgia - 6 years, 6 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.

Calvin Lin Staff - 6 years, 6 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.

Rohit Kumar - 6 years, 6 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'

Krishna Sharma - 6 years, 6 months ago

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

Siddhartha Srivastava - 6 years, 6 months ago

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

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

N should be odd

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

@Calvin Lin – Ok so the problem is making a case x= 0?

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

Sharky Kesa - 6 years, 6 months ago

Good

Ayanlaja Adebola - 6 years, 6 months ago

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

Math	Appears as
Remember to wrap math in `\(` ... `\)` or `\[` ... `\]` to ensure proper formatting.
`2 \times 3`	$2 \times 3$
`2^{34}`	$2^{34}$
`a_{i-1}`	$a_{i-1}$
`\frac{2}{3}`	$\frac{2}{3}$
`\sqrt{2}`	$\sqrt{2}$
`\sum_{i=1}^3`	$\sum_{i=1}^3$
`\sin \theta`	$\sin \theta$
`\boxed{123}`	$\boxed{123}$