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

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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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Good

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

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