9
votes
Show an Inequality $\int_{0}^\infty\frac{\sqrt x\ln(1+x+x^2)}{1+x^2}<(1+\sqrt 3)\pi$
Here is to explicitly evaluate the integral, via $\int_0^\infty \frac{\ln(a^2+x^2)}{b^2+x^2}dx=\frac \pi b\ln(a+b)$
\begin{align}
& \int_{0}^\infty\frac{\sqrt x\ln(1+x+x^2)}{1+x^2}\overset{x\to x^...
- 102k
8
votes
How do I define the solution to an inequality?
I will give an answer along the lines of copper.hat, but going into more detail of how one might arrive at his answer.
We make use of a definition of $\vert t\vert, $ which is:
$
\vert t \vert:=
\...
- 21.2k
5
votes
How do I define the solution to an inequality?
Remember that $A \cap B$ is all elements that belong in both $A$ and $B$.
Is $1$ in both $(1,\infty]$ and $[1,\infty]$? Answer that, and you have your answer.
- 47.5k
5
votes
How do I define the solution to an inequality?
I would suggest plotting the left hand side first to guess the answer.
Let $f(x) = |x-1|+|x+2|$.
For $x \le -2$, $f(x) = 1-x -2-x = -2x-1$, and $-2x-1 > 3 $ iff $-x>2$ iff $x<-2$.
For $x \in [...
- 175k
5
votes
Show an Inequality $\int_{0}^\infty\frac{\sqrt x\ln(1+x+x^2)}{1+x^2}<(1+\sqrt 3)\pi$
$$\int_0^\infty\frac{\sqrt x\ln(1+x+x^2)}{1+x^2}dx=\int_{0}^\infty\frac{\sqrt x\ln x}{1+x^2}dx+\int_{0}^\infty\frac{\ln(x+\frac1x+1)}{\sqrt x(x+\frac1x)}dx=I_1+I_2\tag{0}$$
Using the keyhole contour ...
- 16.7k
4
votes
Prove $\frac{a^2}{(1+ab)^2} + \frac{b^2}{(1+bc)^2} + \frac{c^2}{(1+ac)^2} \geqslant \frac34$ for $a, b, c > 0$ with $abc=1$
Sorry, I cannot comment so it is going to be an answer.
You may want to use the standard substitution: $a=\frac{x}{y},\ b=\frac{y}{z},\ c=\frac{z}{x}$ to obtain equivalently $$\left(\frac{zx}{xy+yz}\...
- 61
4
votes
Tedious Inequality from Spanish Olympiad
Preamble: Here's a standard approach using just AM-GM. There's nothing too tricky here, so I'm surprised the official solution isn't along these lines. Note that all of the ideas are essentially ...
- 71.6k
4
votes
Accepted
Where does the inequality $f(x)\geq (x-11)^2$ comes from?
Since squares of real numbers are always nonnegative,
$$
f(x) = (x-11)^2 + (x^3-3x-1)^2 \ge (x-11)^2 + 0.
$$
- 84.5k
3
votes
How do I define the solution to an inequality?
A faster method is to recall that $|x-x_0|$ measures the distance of the point $x$ from the point $x_0$. So the set of solution of the inequality is given by the points such that $($distance from $1$, ...
- 6,421
3
votes
Accepted
How do I define the solution to an inequality?
As noticed in the comments, at $x=1$ we have
$$|x-1|+ |x+2|=|1-1|+|1+2|=3$$
therefore $x=1$ is not a solution and $S_1$ is the interval $(1,\infty)$.
Similarly we find that $S_2$ is $(-\infty,-2)$.
- 157k
3
votes
Accepted
Showing a function satisfying a certain differential inequality must always be positive
Consider $f(t):=(\cos^2(t)+\sin(2t))/2$. Then $f(0)=1/2$, $f'(0)=1$, and $f$ takes negative values. Now
$$
f'(t)=\frac{1}{2}(-2\cos(t)\sin(t)+ 2\cos(2t)),
$$
$$
f''(t)=\frac{1}{2}(2-4\cos^2(t)-4\sin(...
- 8,144
3
votes
Prove that $\frac {x_1^3}{y_1}+\frac {x_2^3}{y_2}+\ldots+\frac {x_n^3}{y_n}\leq \frac {a^4+b^4}{ab (a^2+b^2)}(x_1^2+x_2^2+\ldots+x_n^2) $
It is reasonable to guess the equality condition happens at $n = 2k$, with $(x_i, y_i ) = (a, b) $ for $k$ values, and $(x_i, y_i) = (b, a) $ for $k$ other values.
OP also observed that $ \frac{ a}{b} ...
- 71.6k
3
votes
If $a^2+b^2+c^2+d^2=4$, what's the maximum value of $a+b+c+d$?
Clearly it is achieved when $a,b,c,d$ are nonnegative. The quadratic mean inequality then gives $$\frac{a+b+c+d}{4}\leq\sqrt{\frac{a^2+b^2+c^2+d^2}{4}}=1.$$
- 42.2k
3
votes
Accepted
Is this unified triangle inequality for sides and angles $\frac{a}{A} + \frac{b}{B} > \frac{c}{C}$ true?
Using law of sines we have
$$\frac{a}{A} + \frac{b}{B} > \frac{c}{C} \iff \frac{\sin A}{A} + \frac{\sin B}{B} > \frac{\sin C}{C}$$
and since for $x\in (0,\pi)$
$$1-\frac x \pi<\frac{\sin x}x&...
- 157k
2
votes
Accepted
Solve the inequality $\frac{2x-1}{x+2} \geq \frac{3x-1}{x+3}$
The problem doesn't make sense in complex numbers, where there is no natural $\le$.
As for the real numbers, there is indeed no solution such that $(x+3)(x+2)>0$, but for $(x+3)(x+2)<0$ your ...
- 41.3k
2
votes
Show an Inequality $\int_{0}^\infty\frac{\sqrt x\ln(1+x+x^2)}{1+x^2}<(1+\sqrt 3)\pi$
Using $x=t^2$ $$I=\frac{\sqrt x\ln(1+x+x^2)}{1+x^2}\,dx=\int\frac{2 t^2 \log\left(t^4+t^2+1\right)}{t^4+1}\,dt$$
$$\frac{2 t^2}{t^4+1}=\frac{2 t^2}{(t^2+i)(t^2-i)}=\frac{1}{t^2+i}+\frac{1}{t^2-i}$$
$$\...
- 269k
2
votes
If $a^2+b^2+c^2+d^2=4$, what's the maximum value of $a+b+c+d$?
Let us write $w:=a-1,\,x:=b-1,\,y:=c-1,\,z:=d-1$. Then maximizing $a+b+c+d$ corresponds to maximizing $w+x+y+z$. Now
$$w+x+y+z=-\tfrac12(w^2+x^2+y^2+z^2).$$
This is maximized just when $w=x=y=z=0$, ...
- 19.6k
2
votes
Prove that $\frac {x_1^3}{y_1}+\frac {x_2^3}{y_2}+\ldots+\frac {x_n^3}{y_n}\leq \frac {a^4+b^4}{ab (a^2+b^2)}(x_1^2+x_2^2+\ldots+x_n^2) $
Alternative proof inspired by Calvin Lin's very nice solution.
Since $\frac{a}{b} \le \frac{x_1}{y_1} \le \frac{b}{a}$, we have
$$\left(\frac{x_1}{y_1} - \frac{a}{b}\right)\left(\frac{x_1}{y_1} - \...
- 41.1k
2
votes
Accepted
Iinequality between spectral norm and the sum of induced $L1$ and $L_\infty$ of a matrix
In the below, denote the induced $p$-norm ($1\le p\le\infty$) of a matrix by $\|\cdot\|_p$. Then
\begin{align*}
\|A\|_2
&=\sqrt{\rho(A^TA)}=\sqrt{\rho(AA^T)}\\
&\le\sqrt{\|AA^T\|_\infty}\quad\...
- 142k
1
vote
Accepted
Proving the monotonicity and unimodality of 2 functions
In this answer, we prove that $z_L(t)$ is unimodal, given $0 < \alpha < \frac45$.
Let
$$f(t) := \frac{\alpha(2t^2+1)-1-\sqrt{(1-\alpha)[\alpha( 8t^3-4t^2-3)+1]+t\alpha^2}}{\alpha(4t^3-1)}.$$
It ...
- 41.1k
1
vote
Solve the inequality $\frac{2x-1}{x+2} \geq \frac{3x-1}{x+3}$
$$\frac{2x-1}{x+2}-\frac{3x-1}{x+3}=-\frac{x^2+1}{(x+2)(x+3)}.$$
So this expression is positive when the denominator is negative (between $-3$ and $-2$).
- 2,908
1
vote
Accepted
How to prove this inequality with parameter $a$?
Please check this example (if I am right):
Let $a = \frac{1}{15}$, and $N = 6$, and
$x_1 = x_2 = x_3 = x_4 = 50$,
and $x_5 = x_6 = 2$. We have
$$\frac{x_1 + x_2 + x_3 + x_4 + x_5 + x_6}{(...
- 41.1k
1
vote
Accepted
Apply Taylor's Theorem to show optimal solution of $f$
Some thoughts.
Remark. Perhaps this answer is not what the OP wants,
since the way to apply Taylor's theorem seems unusual.
Let $g(x_1, x_2, x_3) := f(x_1, x_2, x_3) - x_2x_3$.
Let $\Omega := \{(x_1, ...
- 41.1k
1
vote
Show an Inequality $\int_{0}^\infty\frac{\sqrt x\ln(1+x+x^2)}{1+x^2}<(1+\sqrt 3)\pi$
Feynman’s trick
Via $\sqrt x\mapsto x$ transform the integral into
$$
\int_0^{\infty} \frac{\sqrt{x} \ln \left(1+x+x^2\right)}{1+x^2} d x = 2 \int_0^{\infty} \frac{x^2 \ln \left(1+x^2+x^4\right)}{1+x^...
- 23.4k
1
vote
Show an Inequality $\int_{0}^\infty\frac{\sqrt x\ln(1+x+x^2)}{1+x^2}<(1+\sqrt 3)\pi$
Via $\sqrt x\mapsto x $ transforms the integral into
$$
I=2 \int_0^{\infty} \frac{x^2 \ln \left(1+x^2+x^4\right)}{1+x^4} d x
$$
Via $x\mapsto \frac{1}{x} $ transforms the integral into
$$
I=2 \int_0^...
- 23.4k
1
vote
Is this unified triangle inequality for sides and angles $\frac{a}{A} + \frac{b}{B} > \frac{c}{C}$ true?
Note that by normalizing, we can set $c=1$. By sine law, we have $a=\dfrac{\sin A}{\sin C}$. Therefore, rearranging we have the statement is equivalent to $$\dfrac{\sin A}{A}+\dfrac{\sin B}{B}-\dfrac{\...
- 1,903
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