Articles of maxima minima

a two-variable cyclic power inequality $x^y+y^x>1$ intractable by standard calculus techniques

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

Find max: $\frac{a}{b+2a}+\frac{b}{c+2b}+\frac{c}{a+2c}$

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

Calculus of variations ( interpreting the minimum in first order)

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 […]

How to find $\max\int_{a}^{b}\left (\frac{3}{4}-x-x^2 \right )\,dx$ over all possible values of $a$ and $b$, $(a<b)$?

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 $b$.

Maximum of $x^3+y^3+z^3$ with $x+y+z=3$

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 $\mathbb{R}^+$; then find the minimum value of $\frac{1}{a}+\frac{4}{b}+\frac{9}{c}$ .

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} $ $\Rightarrow \frac{1}{a}+\frac{4}{b}+\frac{9}{c}\ge 6(6^\frac{1}{3}-1)$**

Find a minimum of $x^2+y^2$ under the condition $x^3+3xy+y^3=1$

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 […]

What are the common abbreviation for minimum in equations?

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

A very general method for solving inequalities repaired

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 […]

Lagrange multipliers finding the maximum and minimum.

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 $(0,-1)$ […]