This question already has an answer here: Exponential teaser [closed] 1 answer

This question already has an answer here: Olympiad inequality $\frac{a}{2a + b} + \frac{b}{2b + c} + \frac{c}{2c + a} \leq 1$. 2 answers

I have just begun studying the Calculus of Variations and I have 3 doubts in it. I have written certain things in bold so as anyone who wishes to answer the question but find it quite long can just go through the bold part and he would get an idea as what my doubt is […]

I tried finding the maxima of $f(x)=\frac{3}{4}-x-x^2$ by taking the derivative and so on and use the fact that $\displaystyle\int_{a}^{b}f(x)\,dx \leq M(b-a)$ where $M$ is the global maximum, but then the maximum value depends on the values of $a$ and $b$.

It is given that, $x+y+z=3\quad 0\le x, y, z \le 2$ and we are to maximise $x^3+y^3+z^3$. My attempt : if we define $f(x, y, z) =x^3+y^3 +z^3$ with $x+y+z=3$ it can be shown that, $f(x+z, y, 0)-f(x,y,z)=3xz(x+z)\ge 0$ and thus $f(x, y, z) \le f(x+z, y, 0)$. This implies that $f$ attains it’s maximum […]

If $a+b+c=6$ and $a,b,c$ belongs to positive reals, then find the minimum value of $$\frac{1}{a}+\frac{4}{b}+\frac{9}{c}$$ using AM $\ge HM$ $\frac{a+b+c}{3}\ge\frac{3}{\frac{1}{a}+\frac{1}{b}+\frac{1}{c}}$ ${\frac{1}{a}+\frac{1}{b}+\frac{1}{c}}\ge\frac{3}{2}$ or **why not $AM\ge GM $ $\frac{\frac{1}{a}+\frac{4}{b}+\frac{9}{c}+a+b+c}{6}\ge (\frac{1}{a}\times\frac{4}{b}\times\frac{9}{c}\times a\times b\times c)^\frac{1}{6} $ $\Rightarrow \frac{1}{a}+\frac{4}{b}+\frac{9}{c}\ge 6(6^\frac{1}{3}-1)$**

As in the title, I’ve tried to find a maximum and mininum of $x^2+y^2$ when $x^3+3xy+y^3=1$ holds. It is not too hard to show that $x^2+y^2$ has no maximum, but I can’t find a minimum. Lagrange multiplier gives a dirty calculation so I can’t handle it. Is there any elegant way to find it? Thanks […]

I’m searching for some symbol representing minimum that is commonly used in math equations.

Yesterday, I asked a question about a very general method for solving equations I had found here. As it turned out, there were quite some problems with my method and I got a lot of good feedback. After thinking about it for a while, I’ve come up with a new method. It still uses the […]

Consider the two-variable function $f(x, y) = x^ 2 + 2y^2$ (a) Find the maxima and minima of $f(x, y)$ on the unit circle $x^2 +y^2 = 1$ I have used lagrange multipies to get landa= to $1$ or $2$. Given the points$(1,0),(-1,0),(0,1),(0,-1)$. so the min would be at $(0,0)$ and the max at $(0,-1)$ […]

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