3
votes
Prove $\frac{a+b}{c} + \frac{b+c}{a}+\frac{a+c}{b} + 6 \ge 2\sqrt2(\sqrt\frac{1-a}{a} + \sqrt\frac{1-b}{b} + \sqrt\frac{1-c}{c})$ when $a + b + c = 1$
From a+b+c=1 we get that a+b=1-c etc, so the LHS is equal to $\frac{1-a}{a}+\frac{1-b}{b}+\frac{1-c}{c}+6$. We can use the notation $x=\frac{1-a}{a}, y=\frac{1-b}{b}, z=\frac{1-c}{c}$ and now the ...
- 31
3
votes
Accepted
n complex numbers inside a disk with center $A$ and radius 1
First note that
$$ \tag{1}
\sum_{k=1}^n |z_k| \cdot \sum_{k=1}^n \frac{1}{|z_k|} \ge n^2
$$
by the inequality between harmonic and arithmetic mean.
What we need next is a lower bound of $\left| \...
- 105k
3
votes
Find the min value of $\frac{xy}{z}+\frac{xz}{y}+\frac{yz}{x}$ when ${x^2} + {y^2} + {z^2} = 1$.
We observe that, if $x,y,z>0$ does not hold, then the global minimum does not exist . Therefore, we need the restriction $x,y,z>0$ .
Let $\thinspace\dfrac {xy}{z}=a,\thinspace \dfrac {xz}{y}=b,\...
- 14.1k
3
votes
Combinatorics board olympiad problem Rioplatenes P3
Let's assume that $302$ colors have been used on the board. Since, no rows can admit $5$ (or more) colors, at least $2$ rows must contain exactly $4$ colors so that all the ($4 \times 2$) colors are ...
- 4,280
2
votes
Accepted
Find the minimum positive value that $(f(x)-g(x)+3)$ may attain.
You've made great progress. Using the Arctangent addition formula, along with your result of $\arctan(f(x)) - \arctan(g(x)) = \arctan\left(\frac{1}{2}\right)$, we get that
$$\arctan\left(\frac{f(x) - ...
- 45.8k
2
votes
Accepted
"A secretary had to call all the clients in her company" probably an easy problem that I am really struggling for some reason.
Let the total number of calls be $x$.
Morning: $\frac{1}{3}x$, afternoon: $\frac{3}{5}(x-\frac{1}{3}x)$ (because it is the left over ones), night:$64$.
So we have $x=\frac{1}{3}x+\frac{3}{5}(x-\frac{1}...
- 3,636
2
votes
What is the minimal grid size for which the property stated in this problem still holds?
Let's consider the four color case.
If the square's dimension is $15$, then it contains $15^2 = 225 = 4\cdot 56+1$ cells and at least $57$ of these must be of the same color, say red.
Of these $57$ ...
- 1,711
2
votes
Inequality with complex numbers with the same modulus
This solution is almost surely not the best, but it works.
$\color{#FF6200}{\bf{Lemma.}}$ If $\lambda\in\mathbb{R}$, then $\forall X\geq4\lambda:Q_\lambda(X):=X^2-\lambda(2+\lambda)X+\lambda^2(1-\...
- 1,343
1
vote
Find the minimum positive value that $(f(x)-g(x)+3)$ may attain.
In what follows, $f(x)$ and $g(x)$ will be written as $f$ and $g$ respectively.
Note that
$$\tan^{-1} f = \tan^{-1} g + \tan^{-1} \left( \frac{1}{2} \right) $$
$$\implies f = \frac{g+ \frac{1}{2}}{...
- 2,728
1
vote
Accepted
Prove $\sum\limits_{\mathrm{cyc}} \sqrt{5a+5b+8ab}\ge 3\sqrt{2}+2\sqrt{5}$ for $ab+bc+ca=1$
A proof using Holder inequality.
By Holder inequality, we have
\begin{align*}
&\left(\sum_{\mathrm{cyc}} \sqrt{5a + 5b + 8ab}\right)^2
\sum_{\mathrm{cyc}} (5a + 5b + 8ab)^2\Big(5a + 5b + (6\sqrt{...
- 32.9k
1
vote
Prove that $\sum\limits_{cyc}\sqrt{\frac{8ab+8ac+9bc}{(2b+c)(b+2c)}}\geq5$
My second proof.
In Michael Rozenberg's answer, Nguyenhuyen_AG's idea is the so-called isolated fudging.
I got the same form using the isolated fudging technique.
We need to prove that
$$\sqrt{\frac{...
- 32.9k
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