What is the maximum value of $x^2+y^2$, where $(x,y)$ are solutions to:  $$2x^2+5xy+3y^2=2$$ and   $$6x^2+8xy+4y^2=3$$  Note: Calculus is not allowed. I tried everything I could but whenever I got for example or $x^2+y^2=f(y)$ or $f(x)$ the function $f$ would always be a concave up parabola, so I could not find a maximum for either variable. And by the way I know that you can solve for $x$ and $y$ using the quadratic formula and get $4$ different solutions but I am looking for less messy way. I've asked this question , but I didn't get the nice answer I wanted. Thanks.  This question came from a math competition from the Math Honor Society, Mu Alpha Theta.
122667
What is the maximum value of $x^2+y^2$, where $(x,y)$ are solutions to $2x^2+5xy+3y^2=2$ and $6x^2+8xy+4y^2=3$. (calculus is not allowed). I tried everything I could but whenever I got for example $or$ $x^2+y^2=f(y)$ or $f(x)$ the function $f$ would always be a concave up parabola, so I could not find a maximum for either variable. However, I also don't see how you could solve it if you leave both variables on one side. And by the way I know that you can solve for $x$ and $y$ using the quadratic formula and get $4$ different solutions but I am looking for a much more efficient way than that.   This question came from a math competition from the Math Honor Society, Mu Alpha Theta.
121648
Let $f$ be a polynomial of degree $n$ at most $3$ such that there exists some $\alpha,\beta$ satisfying $\int_0^2 f(x)dx=f(\alpha) +f(\beta)$ for all such $f$. Find the value of $3(\alpha-\beta)^2$  Attempt: We need to find the value of $3(\alpha-\beta)^2$. Now, this expression reminds of the function $f(x) = (x-\beta)^3$. Differentiating this function at $x=\alpha$ will give us the desired result.  Hence, we just need to know the derivative of $k(x)=(x-\beta)^3$ at every point.  However, I am unable to move forward with this intuition. Could anyone please help me to move ahead in this problem.  Thank you very much for your help in this regard.
12728
Maximum value of $\cos \alpha_{1}\cdot \cos \alpha_{2}\cdot \cos \alpha_{3}\cdot \cos \alpha_{4}....\cos \alpha_{n}.$ If it is given that   $\cot \alpha_{1}\cdot \cot \alpha_{2}\cdot \cot \alpha_{3}.......\cot \alpha_{n}  = 1$ and $\displaystyle 0 \leq \alpha_{1},\alpha_{2},\alpha_{3},.......,\alpha_{n}\leq \frac{\pi}{2}.$  $\bf{My\; Try::}$ Using $\bf{A.M\geq G.M}$ $$\displaystyle \frac{\sin^2 \alpha_{1}+\sin^2 \alpha_{2}+\sin^2 \alpha_{3}+........+\sin^2 \alpha_{n}}{n}\geq \sqrt[n]{\sin^2 \alpha_{1}\cdot \sin^2 \alpha_{2}...\sin^2 \alpha_{n}}$$ So we get $$\displaystyle \sin^2 \alpha_{1}+\sin^2 \alpha_{2}+\sin^2 \alpha_{3}+........+\sin^2 \alpha_{n}\geq n\cdot \sqrt[n]{\sin^2 \alpha_{1}\cdot \sin^2 \alpha_{2}...\sin^2 \alpha_{n}}.......(1)$$ Similarly $$\displaystyle \frac{\cos^2 \alpha_{1}+\cos^2 \alpha_{2}+\cos^2 \alpha_{3}+........+\cos^2 \alpha_{n}}{n}\geq \sqrt[n]{\cos^2 \alpha_{1}\cdot \cos^2 \alpha_{2}...\cos^2 \alpha_{n}}$$ So we get $$\displaystyle \cos^2 \alpha_{1}+\cos^2 \alpha_{2}+\cos^2 \alpha_{3}+........+\cos^2 \alpha_{n}\geq n\cdot \sqrt[n]{\cos^2 \alpha_{1}\cdot \cos^2 \alpha_{2}...\cos^2 \alpha_{n}}.......(2)$$ Now Adding $(1)$ and $(2)\;,$ We get $$n\geq 2n\cdot \sqrt[n]{\cos^2 \alpha_{1}\cdot \cos^2 \alpha_{2}...\cos^2 \alpha_{n}}$$ Because above it is given that $$\sin \alpha_{1}\cdot \sin\alpha_{2}\cdot\sin \alpha_{3}.........\sin\alpha_{n}=\cos\alpha_{1} \cdot\cos\alpha_{2}\cdot \sin\alpha_{3}......\cos\alpha_{n}$$ So we get $$\displaystyle \cos \alpha_{1}\cdot \cos \alpha_{2}\cdot \cos \alpha_{3}\cdot \cos \alpha_{4}....\cos \alpha_{n}\leq \frac{1}{2^{\frac{n}{2}}}$$ My question is that can we solve it using any other short method? If yes then please explain it here. Thanks.
111075
Problem Statement  Let $f(x) = x^3 + ax^2 + bx + c$ and $g(x) = x^3 + bx^2 + cx + a$ where $a,b,c$ are integers with $c\not=0$   Suppose that the following conditions hold:   $f(1)=0$ the roots of $g(x)=0$ are the squares of the roots of $f(x)=0$      $$\text{Find the value of} \: \: a^{2013} + b^{2013} + c^{2013}$$   My attempt  Let $f(x) = (x-p)(x-q)(x-r) \: \:$ and $g(x) = (x - p^2)(x-q^2)(x-r^2)$  Expanding we get $$f(x) = x^3 - (p + q + r)x^2 + (pq + pr + qr)x - pqr$$  $$g(x) = x^3 - (p^2 + q^2 + r^2)x^2 + (p^2q^2 + p^2r^2 + r^2q^2)x - p^2q^2r^2$$  And so we have $$a = -(p+q+r) = -p^2q^2r^2$$ $$b = pq + pr + qr = p^2 + q^2 + r^2$$ $$c = -pqr = p^2q^2 + p^2r^2 + r^2q^2$$  Then we get the following relations  $$a = -c^2$$ $$b = a^2 - 2b \implies 3b = a^2 \implies b = \frac{c^4}{3}$$  From $f(1) = 0$ we have $a + b + c = -1 \implies \dfrac{c^4}{3} + c - c^2 = -1$  However, upon graphing this function, I saw that there are no integer solutions for $c$. Have I done something wrong?  Edit  For future readers, my approach can simplified alot if you replace $r$ with $1$ since $f(1) = g(1) = 0$
118396
How do I find the maximum value for $x\cos {λ}+y\sin {λ}$ for fixed $x,y$ and varying $λ$ ? I would appreciate if only hint is given.
105982
Find all $x \in (-1, +\infty)$ such that $(x^2 + 4x + 3)^x + (2x + 4)^x = (x^2 + 4x + 5)^x$.  What I have done so far was a substitution $y = x + 2$ which results in a nicer form of the equation:   $(y - 1)^{y - 2}(y + 1)^{y - 2} + 2^{y - 2}y^{y - 2} = (y^2 + 1)^{y - 2}$ and $y &gt; 1$, which can then be rewritten as:  $$ \left( \frac{y^2 + 1}{2y} \right)^{y - 2} - \left( \frac{y^2 - 1}{2y} \right)^{y - 2} = 1 $$ Also, my intuition is that $y = 4$ is the unique solution and initially I wanted to prove that this function of $y$ is injective, but it didn't work either.  Do you have any suggestion on how I can continue?
92099
Find all real solutions for $x$ in $2(2^x−1)x^2+(2^{x^2}−2)x=2^{x+1}−2$.   I started off by dividing $2(2^x- 1) x^2 + (2^{x^2}-2)x = 2^{x+1} -2$ by $2$, and I got $(2^x-1)x^2 + (2^{x^2-1}-1)x = 2^x -1$. I tried dividing by $2^x-1$ on both sides which would give me a simple quadratic to solve for $x$, but I don't know how to simplify $\frac{(2^{x^2}-1)}{(2^x-1)}$. Am I missing a simple trick here, or am I on a completely wrong path to solving this?
47332
Find the global maxima/minima of $f(x,y,z) = x+y+z$ for points inside of $A = \{ (x,y,z) \in \mathbb{R}^3: x^2-y^2 = 1 \wedge 2x+z = 1 \}$   I renamed the conditions of $A$ to a function $g(x,y,z) = x^2-y^2-2x-z = 0$ in order to be able to use Lagrange multipliers.  Deriving $f$: $\nabla(x,y,z) = (1,1,1)$  Deriving $g$:   $g_x = 2x-2$ $g_y = -2y$ $g_z = -1$   By solving the Lagrange system:  \begin{cases} 1 = \lambda(2x-2) \\ 1 = \lambda (-2y) \\ 1 = \lambda (-1) \\ g(x,y) = 0 \end{cases}  I get that $(x,y,z) = (\frac{1}{2}, \frac{1}{2}, -1)$ for $\lambda = -1$. But that point is not inside A.  Is this enough to guarantee that $f|_A$ does not have either maxima nor minima?  (Since the second derivatives are all zero, the second derivative test doesn't give any information.)  Thanks!  EDIT:  By parametrizing $y = \pm \sqrt{x^2-1}$ and $z=1-2x$ and doing the composition with $f$ I get that $f(x, \sqrt{x^2-1}, 1-2x)$ is monotonically increasing and $f(x, -\sqrt{x^2-1}, 1-2x)$ monotonically decreasing. Would that prove that $f$ never reaches either maxima nor minima?
84034
I need help solving this past exam question, the professor posted exam questions with out solutions to last year's exam, I tried reduction of order, however that seemed to complicate things. And then I tried Euler-Cauchy method however I realized that won't work due to the $(2x^2 + 1)y''$.  Given $y_1=x$ is a solution $$(2x^2 + 1)y'' − 4xy' + 4y = 0$$ Find the second solution $y_2$ for equation.
102624
The question is : prove that $y^2 = x^3+(x+4)^2$ have no solutions in positive integers $x,y$.  I tried to play with the equation and get to $ x^3 = (y+x+4)(y-x-4)$, if $p|x$ then $p^3 | x^3$ and $ p | y+x+4$ or $ p | y-x-4$ , assume that $p$ divides both then we get that $ p | 2x-8$ and since $p|x$ we get that $ p | 8$ meaning $p=2$, so if we assume that $x$ is odd then $ p^3 | y+x+4$ or $p^3| y-x-4$ but not both and so we can write $y+x+4 = b^3 $ and $ y-x-4= a^3$, but that did not help me solving the problem.
42133
Demand $z=x+y$ and $x^2/4 + y^2/5 + z^2/25 = 1$. What is the maximum value of $f(x,y,z) = x^2+y^2+z^2$?   I've been attempting this with Lagrange multipliers in a few different ways. However, the resulting system of equations with two lagrangians has so many variables that it becomes very complicated. Can someone show how this is to be done manually?  I also attempted turning it into two unknowns by replacing $z$ with $x+y$. However, this also led nowhere.
54751
Let $f\in C^2(\mathbb{R},\mathbb{R})$.   Assume there exist  $(a,b) \in \mathbb{R}^2$ such that  $\forall x \in \mathbb{R}, (f(x))^{2} \leqslant a$ and $(f'(x))^{2} + (f''(x))^{2} \leqslant b$.   Prove that: $$(f(x))^{2} + (f'(x))^{2} \leqslant \max(a,b)$$ for all $x \in \mathbb{R}$.   I tried to use taylor lagrange with the integral form of the remainder but I did not succeed.  Ì really don't know how to tackle  the problem.   Any suggestion, Hint (or Answer) will be very helpful,  Thank you in advance,
16282
How to solve this equation $$x^{2}=2^{x}$$   where $x \in \mathbb{R}$.  Por tentativa erro consegui descobri que $2$ é uma solução, mas não encontrei um método pra isso. Alguma sugestão?(*)  (Translation: By trying different values I've found that $2$ is a solution, but I couldn't find any method to this though. Any suggestions? )
7574
Find all integer solutions to $x^2+4=y^3$.  Some obvious solutions are $(x,y)=(\pm2,2)$. Are these the only ones?
107328
a) How does $x^2y''-3xy'+3y=0$ can be solved? I know how to solve for constant coefficients, but in this case they are functions...  b) In which maximum interval there is a solution that  confirms $y'(1)=6, y(1)=4$?
52355
For $a,b,c,d \in \mathbb{R}$, is $\sqrt{a^2+b^2} + \sqrt{c^2+d^2} \geq \sqrt{(a+c)^2 + (b+d)^2}$? I have tried to show this using the fact that $\sqrt{x+y}\leq\sqrt{x}+\sqrt{y}\; $  for non-negative $x,y$ but haven't been able to arrive at a proof.
101787
Let $P(x)=4x^2+6x+4$ and $Q(y)=4y^2-12y+25$.  Find the unique pair of real numbers $(x,y)$ that satisfy $P(x)Q(y)=28$    I can solve this question graphically. $$P(x)Q(y)=28\implies(4x^2+6x+4)(4y^2-12y+25)=28$$ $$4x^2+6x+4=\frac{28}{4y^2-12y+25}$$ I drew the graph of $4x^2+6x+4$ and found it is a upward parabola with minimum value at $(-\frac{3}{4},\frac{7}{4})$ and I drew the graph of $\frac{28}{4x^2-12x+25}$. I found that it is a bell shaped curve whose maximum value occurs at $(\frac{1}{4},\frac{7}{4})$. So I found that the common abscissa is $y=\frac{7}{4}$ which occurs at $x=-\frac{3}{4}$ for the first curve and at $x=\frac{1}{4}$ for the second curve, so solution should be $(-\tfrac{3}{4},\tfrac{1}{4})$  But I do not  know how to solve it algebraically.