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

a+bc2+a+cb2+b+ca22(1a+1b+1c)\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 aa, bb and cc.


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

f(x2y2)=x2(f(y))2f(x^2 - y^2) = x^2 - (f(y))^2

for all real numbers xx and yy.


3) Consider the polynomial p(x)=x33x+1p(x) = x^3 - 3x + 1

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

(b) If the solutions to p(x)=0p(x) = 0 are α\alpha, β\beta and γ\gamma, find a monic cubic polynomial whose roots are 1α2\dfrac {1}{\alpha^2}, 1β2\dfrac {1}{\beta^2} and 1γ2\dfrac {1}{\gamma^2}.


4) The equation x4+ax3+bx2+cx+d=0x^4 + ax^3 + bx^2 + cx + d = 0 has four real positive roots.

Prove that

(a) ac16dac \geq 16d

(b) b236db^2 \geq 36d


5) A sequence of real functions fnf_n is defined by

f1(x)=x2+48f_1(x) = \sqrt{x^2 + 48}

and

fn+1(x)=x2+6fn(x)f_{n + 1}(x) = \sqrt{x^2 + 6 f_n(x)}

for n1n \geq 1.

Find all real solutions of the equation f2007(x)=2xf_{2007}(x) = 2x


6) Let nn be a positive integer. Suppose that a1,a2,,an,b1,b2,,bna_1, a_2, \ldots, a_n, b_1, b_2, \ldots, b_n are positive real numbers such that

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

Prove that

i=1nai2ai+bi12i=1nai\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
4 years, 6 months ago

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

  2. Let P(x,y)P(x,y) denote the equation. Then P(0,0)    f(0){0,1}P(0,0)\implies f(0)\in\{0,-1\}. If f(0)=0f(0)=0 then for all non negative xx we have P(x,0)    f(x)=xP(\sqrt x,0)\implies f(x)=x. Now P(0,y)    f(y)=yP(0,\sqrt y)\implies f(-y)=-y for all non negative yy. Thus f(x)=xf(x)=x is a solution for all xx, which clearly works. If f(0)=1f(0)=-1 then P(x,0)    x1P(\sqrt x,0)\implies x-1 for all non negative xx. But choosing positive reals x>yx>y we see that this doesn't satisfy P(x,y)P(x,y). So only solution is f(x)=x xRf(x)=x~\forall x\in\mathbb R.

  3. The discriminant of the given cubic is 81>081>0, so the roots are all real and distinct. Let our desired cubic be q(x)=x3px2+qxrq(x)=x^3-px^2+qx-r. By Vieta's we must have p=α2β2+β2γ2+γ2α2p=\alpha^2\beta^2+\beta^2\gamma^2+\gamma^2\alpha^2, q=α2+β2+γ2q=\alpha^2+\beta^2+\gamma^2 and r=1r=1. Using the Vieta equations α+β+γ=0\alpha+\beta+\gamma=0 and αβ+βγ+γα=3\alpha\beta+\beta\gamma+\gamma\alpha=-3 we can easily find α2+β2+γ2=6\alpha^2+\beta^2+\gamma^2=6 and α2β2+β2γ2+γ2α2=9\alpha^2\beta^2+\beta^2\gamma^2+\gamma^2\alpha^2=9. So q(x)=x39x2+6x1q(x)=x^3-9x^2+6x-1.

  4. If the roots are p,q,r,sp,q,r,s, by Vieta's ac16d  (p+q+r+s)(pqr+pqs+prs+qrs)16pqrsac\ge 16d~\Leftrightarrow ~(p+q+r+s)(pqr+pqs+prs+qrs)\ge 16pqrs which follows by applying AM-GM on each bracket. Also b236d (pq+pr+ps+qr+qs+rs)236pqrsb^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 f1(x)=2xf_1(x)=2x has only solution x=4x=4 which, by induction, is the root of all fn(x)2xf_n(x)-2x. It easily follows that f1(x)2xf_1(x)-2x is monotonically decreasing. So by induction all of fn(x)2xf_n(x)-2x, being compositions of monotone mappings, are monotonically decreasing. So each fn(x)2xf_n(x)-2x has only one root, which is 44.

  6. Direct application of Titu's Lemma. We have ai2ai+bi(ai)2ai+bi=(ai)22ai=12ai.\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 - 4 years, 6 months ago

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How you calculated discriminant of a cubic please help me

Aman Sharma - 4 years, 6 months ago

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

Jubayer Nirjhor - 4 years, 6 months ago

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The last one is quite easy.
(i=1nai2ai+bi)(i=1n(ai+bi))(i=1nai)2\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

(i=1nai2ai+bi)((i=1nai)22i=1nai)i=1nai2ai+bi12i=1nai\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 - 4 years, 6 months ago

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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 ai2bi2ai+bi=(aibi)=0 \sum \frac{ a_i ^2 - b_i ^ 2 } { a_i + b_i} = \sum (a_i - b_i) = 0 .
Hence, this gives us ai2ai+bi=bi2ai+bi \sum \frac{ a_i ^ 2 } {a_i + b_i } = \sum \frac { b_i^2 } { a_i + b_i } .
We want to show that ai2+bi2ai+bi=2ai2ai+biai \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 ai2+bi22ai+bi2 \sqrt{ \frac{ a_i ^2 + b_i^2 } { 2} } \geq \frac{ a_i + b_i} { 2} , or that ai2+bi2ai+biai+bi2 \frac{ a_i^2 + b_i^2} {a_i + b_i } \geq \frac{ a_i + b_i} {2} .

Hence, we get that ai2+bi2ai+biai+bi2=ai \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 - 4 years, 6 months ago

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Solution to the first problem -

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

(a+b+c)3(1a2+1b2+1c2)>(1a+1b+1c)\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 a+b+c3>3(1a+1b+1c)\frac{a + b + c}{3} > \frac{3}{(\frac{1}{a} + \frac{1}{b} +\frac{1}{c})}

the problem reduces to proving -

3(1a2+1b2+1c2)>(1a+1b+1c)23(\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 - 4 years, 6 months ago

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For 2nd problem I guess there are many

f(x2y2)=x2+(f(y))2\displaystyle f(x^{2} - y^{2}) = x^{2} + (f(y))^{2}

Put x = 0

f(y2)=(f(y))2\displaystyle f(-y^{2}) = (f(y))^{2}

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

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

There can be \infty functions for different 'n'

Krishna Sharma - 4 years, 6 months ago

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

Siddhartha Srivastava - 4 years, 6 months ago

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

Krishna Sharma - 4 years, 6 months ago

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f(x)=x2 f(x) = - x^2 does not satisfy the original functional equation.

Calvin Lin Staff - 4 years, 6 months ago

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N should be odd

Krishna Sharma - 4 years, 6 months ago

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@Krishna Sharma Similarly, f(x)=x3 f(x) = - x^3 does not satisfy the original functional equation.

Calvin Lin Staff - 4 years, 6 months ago

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

Krishna Sharma - 4 years, 6 months ago

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I think you did something wrong. It asks for f(x2y2)=x2(f(y))2f(x^2 - y^2) = x^2 - (f(y))^2.

Sharky Kesa - 4 years, 6 months ago

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Good

Ayanlaja Adebola - 4 years, 6 months ago

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

Sharky Kesa - 4 years, 6 months ago

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