inequality (1 Viewer)

s97127 · Mar 24, 2021

a, b, c > 0 and ab+bc+ca = 1

Prove:

pikachu975 · Mar 24, 2021

This is probably gonna be a dumb question but what's the symbol on the LHS, e.g. (something - a^2)?

s97127 · Mar 24, 2021

pikachu975 said:
This is probably gonna be a dumb question but what's the symbol on the LHS, e.g. (something - a^2)?

(1-a^2)/(1+a^2) + (1-b^2)/(1+b^2) + 2/sqrt(1+c^2) <= 2*sqrt(2)

fan96 · Mar 25, 2021

First rearrange the inequality:

$\frac{1-a^2}{1+a^2} + \frac{1-b^2}{1+b^2} \le 2 \sqrt 2 - \frac 2 {\sqrt{1+c^2}}$

$\iff 1 - \frac{2a^2}{1+a^2} + 1 - \frac{2b^2}{1+b^2} \le 2 \sqrt 2 - \frac 2 {\sqrt{1+c^2}}$

$\iff 1 -\frac{a^2+b^2+2a^2b^2}{(1+a^2)(1+b^2)} \le \sqrt 2 - \frac 1 {\sqrt{1+c^2}}$

$\iff \frac{1 - a^2b^2}{(1+a^2)(1+b^2)} + \frac 1 {\sqrt{1+c^2}} \le \sqrt 2.$

Because $ab + ac + bc = 1$ , we have

$c = \frac{1-ab}{a+b}$

$\implies \sqrt{1+c^2} = \frac{\sqrt{(1+a^2)(1+b^2)}}{a+b}.$

Importantly, because $c > 0$ , this explicit formula for $c$ shows that $ab < 1$ and thus $a,b < 1$ .

Therefore the inequality is equivalent to

$\frac{1 - a^2b^2}{(1+a^2)(1+b^2)} + \frac {a+b}{\sqrt{(1+a^2)(1+b^2)}} \le \sqrt 2.$

Finally,

$\text{LHS} < \frac{1 - (1)^2(1)^2}{(1+0^2)(1+0^2)} + \frac{1+1}{\sqrt{(1+0^2)(1+0^2)}} = \sqrt 2.$

CM_Tutor · Mar 26, 2021

fan96 said:
Importantly, because $c > 0$ , this explicit formula for $c$ shows that $ab < 1$ and thus $a,b < 1$ .

Therefore the inequality is equivalent to

$\frac{1 - a^2b^2}{(1+a^2)(1+b^2)} + \frac {a+b}{\sqrt{(1+a^2)(1+b^2)}} \le \sqrt 2.$

Finally,

$\text{LHS} < \frac{1 - (1)^2(1)^2}{(1+0^2)(1+0^2)} + \frac{1+1}{\sqrt{(1+0^2)(1+0^2)}} = \sqrt 2.$

I am not certain that the last part of this is valid.

I agree that we have $ab < 1$ and that $0 < a, b < 1$ , but as a consequence the term

$T_1 = \frac{1 - a^2b^2}{(1+a^2)(1+b^2)}$

must be non-zero:

$\begin{align*} 0 &< ab < 1 \\ 0^2 &< a^2b^2 < 1^2 = 1 \\ 0 &> - a^2b^2 > -1 \\ \text{So,} \qquad 1 &> 1 - a^2b^2 > 0 \end{align*}$

$\begin{align*} 0 &< a < 1 \\ 0 &< a^2 < 1^2 = 1 \\ 1 &< 1 + a^2 < 2 \qquad \qquad \qquad \text{and similarly for $b^2$} \\ \text{So,} \qquad 1 &< (1 + a^2)(1 + b^2) < 4 \end{align*}$

$\text{So,}\ T_1 = \frac{\text{positive value between 0 and 1}}{\text{positive value between 1 and 4}} \qquad \implies \qquad 0 < T_1 < 1$

Similar reasoning allows us to conclude that

$T_2 = \frac {a+b}{\sqrt{(1+a^2)(1+b^2)}}$

satisfies

$0 < T_2 < \sqrt{2}$

And, while it may be true that $0 < T_1 + T_2 < \sqrt{2}$ , this evidence does not support any conclusion stronger than $0 < T_1 + T_2 < 1+\sqrt{2}$

Consider some values:

Case 1: Both near zero

$\text{Take } a = b = \frac{1}{10}$

$\begin{align*} T_1 &= \frac{1 - a^2b^2}{(1+a^2)(1+b^2)} = \frac{1 - \left(\frac{1}{10}\right)^2\left(\frac{1}{10}\right)^2}{\left(1+\left(\frac{1}{10}\right)^2\right)\left(1+\left(\frac{1}{10}\right)^2\right)} \\ &= \frac{1 - \frac{1}{10000}}{\left(1+\frac{1}{100}\right)\left(1+\frac{1}{100}\right)} = \frac{\frac{9999}{10000}}{\left(\frac{101}{100}\right)^2} = \frac{9999}{10201} \end{align*}$

$\begin{align*} T_2 &= \frac{a+b}{\sqrt{(1+a^2)(1+b^2)}} = \frac{\frac{1}{10}+\frac{1}{10}}{\sqrt{\left(1+\left(\frac{1}{10}\right)^2\right)\left(1+\left(\frac{1}{10}\right)^2\right)}} \\ &= \frac{\frac{2}{10}}{\sqrt{\left(\frac{101}{100}\right)^2}} = \frac{\frac{2}{10}}{\frac{101}{100}} = \frac{20}{101} \end{align*}$

$\text{So, } T_1 + T_2 = \frac{9999}{10201} + \frac{20}{101} = \frac{9999 + 2020}{10201} = 1\frac{18}{101} = 1.1782... < \sqrt{2}$

Case 2: Both near one

$\text{Take } a = b = \frac{9}{10}$

$\begin{align*} T_1 &= \frac{1 - a^2b^2}{(1+a^2)(1+b^2)} = \frac{1 - \left(\frac{9}{10}\right)^2\left(\frac{9}{10}\right)^2}{\left(1+\left(\frac{9}{10}\right)^2\right)\left(1+\left(\frac{9}{10}\right)^2\right)} \\ &= \frac{1 - \frac{6561}{10000}}{\left(1+\frac{81}{100}\right)\left(1+\frac{81}{100}\right)} = \frac{\frac{3439}{10000}}{\left(\frac{181}{100}\right)^2} = \frac{3439}{32761} \end{align*}$

$\begin{align*} T_2 &= \frac{a+b}{\sqrt{(1+a^2)(1+b^2)}} = \frac{\frac{9}{10}+\frac{9}{10}}{\sqrt{\left(1+\left(\frac{9}{10}\right)^2\right)\left(1+\left(\frac{9}{10}\right)^2\right)}} \\ &= \frac{\frac{18}{10}}{\sqrt{\left(\frac{181}{100}\right)^2}} = \frac{\frac{18}{10}}{\frac{181}{100}} = \frac{180}{181} \end{align*}$

$\text{So, } T_1 + T_2 = \frac{3439}{32761} + \frac{180}{181} = \frac{3439 + 32580}{32761} = 1\frac{18}{181} = 1.099447... < \sqrt{2}$

Case 3: Both near middle

$\text{Take } a = b = \frac{1}{2}$

$\begin{align*} T_1 &= \frac{1 - a^2b^2}{(1+a^2)(1+b^2)} = \frac{1 - \left(\frac{1}{2}\right)^2\left(\frac{1}{2}\right)^2}{\left(1+\left(\frac{1}{2}\right)^2\right)\left(1+\left(\frac{1}{2}\right)^2\right)} \\ &= \frac{1 - \frac{1}{16}}{\left(1+\frac{1}{4}\right)\left(1+\frac{1}{4}\right)} = \frac{\frac{15}{16}}{\left(\frac{5}{4}\right)^2} = \frac{3}{5} \end{align*}$

$\begin{align*} T_2 &= \frac{a+b}{\sqrt{(1+a^2)(1+b^2)}} = \frac{\frac{1}{2}+\frac{1}{2}}{\sqrt{\left(1+\left(\frac{1}{2}\right)^2\right)\left(1+\left(\frac{1}{2}\right)^2\right)}} \\ &= \frac{1}{\sqrt{\left(\frac{5}{4}\right)^2}} = \frac{1}{\frac{5}{4}} = \frac{4}{5} \end{align*}$

$\text{So, } T_1 + T_2 = \frac{3}{5} + \frac{4}{5} = 1\frac{2}{5} = 1.4 < \sqrt{2} = 1.4142...$

Case 4: One near zero, one near one

$\text{Take } a = \frac{1}{10} \text{ and } b = \frac{9}{10}$

$\begin{align*} T_1 &= \frac{1 - a^2b^2}{(1+a^2)(1+b^2)} = \frac{1 - \left(\frac{1}{10}\right)^2\left(\frac{9}{10}\right)^2}{\left(1+\left(\frac{1}{10}\right)^2\right)\left(1+\left(\frac{9}{10}\right)^2\right)} \\ &= \frac{1 - \frac{81}{10000}}{\left(1+\frac{1}{100}\right)\left(1+\frac{81}{100}\right)} = \frac{\frac{9919}{10000}}{\frac{101 \times 181}{100^2}} = \frac{9919}{18281} \end{align*}$

$\begin{align*} T_2 &= \frac{a+b}{\sqrt{(1+a^2)(1+b^2)}} = \frac{\frac{1}{10}+\frac{9}{10}}{\sqrt{\left(1+\left(\frac{1}{10}\right)^2\right)\left(1+\left(\frac{9}{10}\right)^2\right)}} \\ &= \frac{1}{\sqrt{\frac{101 \times 181}{100^2}}} = \frac{1}{\sqrt{\frac{18281}{100^2}}} = \frac{100}{\sqrt{18281}} \end{align*}$

$\text{So, } T_1 + T_2 = \frac{9919}{18281} + \frac{100}{\sqrt{18281}} = 1.28219... < \sqrt{2}$

Looking at these four cases, we see that $T_1$ is typically not close to zero, and that the sum does stay below $\sqrt{2}$ . The conclusion based on $T_1 \approx 0$ and $T_2 \approx \sqrt{2}$ is not reflected in any of these cases, however.

fan96 · Mar 26, 2021

My bad, I didn't check that last step correctly.

This may not quite be within the MX2 syllabus, but if you differentiate the expression

$\frac{1 - a^2b^2}{(1+a^2)(1+b^2)} + \frac {a+b}{\sqrt{(1+a^2)(1+b^2)}}$

twice, as a function of $a$ and then as a function of $b$ , it can be shown that there is exactly one point in the unit square $(\sqrt 2 - 1, \sqrt 2 - 1)$ that is a stationary point in both cases.

(The derivative actually only needs to be taken once, as $a$ and $b$ are symmetric in the expression.)

This point can be shown to be a local maximum by analysing the two derivatives.

The function attains the value of $\sqrt 2$ at this point, so $\sqrt 2$ must be its maximum value over the unit square. (It's actually a global maximum, but that isn't necessary to show.)

CM_Tutor · Mar 28, 2021

One slight cheat would be to say that the equation that we have is symmetrical in $a$ and $b$ and thus that the max / min likely lies on the diagonal $y=x$ . We would then need to examine the function:

$F= \frac{1-x^4}{\left(1+x^2\right)^2} + \frac{2x}{1+x^2}$

in the domain $0 < x < 1$ . We would then find:

$\frac{dF}{dx} = \frac{-2\left(x^2+2x-1\right)}{\left(x^2+1\right)^2}$

which is zero when $(x+1)^2 = 2$ and so the only stationary point in the domain corresponds to $x = a = b =\sqrt{2} - 1$ , as @fan96 stated.

$\begin{align*} \frac{d^2F}{dx^2} &= \frac{4\left(x^3 + 3x^2 - 3x - 1\right)}{\left(x^2+1\right)^3} \\ \text{At $x = \sqrt{2} - 1$} \qquad \frac{d^2F}{dx^2} &= \frac{4\left[\left(\sqrt{2}\right)^3 -3\left(\sqrt{2}\right)^2 + 3\left(\sqrt{2}\right)^1 - 1 + 3\left(3 - 2\sqrt{2}\right) - 3\left(\sqrt{2} - 1\right) - 1\right]}{\left(3 - 2\sqrt{2} +1\right)^3} \\ &= \frac{4\left(2\sqrt{2} - 6 + 3\sqrt{2} - 1 + 9 - 6\sqrt{2} - 3\sqrt{2} + 3 - 1\right)}{\left(4 - 2\sqrt{2}\right)^3} \\ &= \frac{4\left(4 - 4\sqrt{2}\right)}{\left[2\left(2 - \sqrt{2}\right)\right]^3} \\ &= \frac{16\left(1 - \sqrt{2}\right)}{8\left[2^3 - 3(2)^2\sqrt{2} + 3(2)\left(\sqrt{2}\right)^2 - \left(\sqrt{2}\right)^3\right]} \\ &= \frac{2\left(1 - \sqrt{2}\right)}{8 - 12\sqrt{2} + 12 - 2\sqrt{2}} \\ &= \frac{2\left(1 - \sqrt{2}\right)}{2\left(10 - 7\sqrt{2}\right)} \times \frac{10+7\sqrt{2}}{10+7\sqrt{2}} \\ &= \frac{10+7\sqrt{2} - 10\sqrt{2}-7 \times 2}{10^2-\left(7\sqrt{2}\right)^2} \\ &= -\frac{4 + 3\sqrt{2}}{2} \end{align*}$

Proving that $x=a=b=\sqrt{2}-1$ corresponds to a maximum.

Substituting $x=\sqrt{2}-1$ will show that $F_{\text{max}} = \sqrt{2}$ and thus completes the proof of the inequality.

I don't plan to type it out, but I have checked, and when $x=\sqrt{2}-1$ , we get

$\frac{1-x^4}{\left(1+x^2\right)^2} = \frac{\sqrt{2}}{2}$

and

$\frac{2x}{1+x^2} = \frac{\sqrt{2}}{2}$

and thus get

$F_{\text{max}} = \frac{\sqrt{2}}{2} + \frac{\sqrt{2}}{2} = \sqrt{2}$

s97127 · Mar 28, 2021

idkkdi · Mar 29, 2021

s97127 said:
View attachment 30464

that is some shoddy writing lol

Paradoxica · Mar 29, 2021

Random observation I want to make. Let

$a = \tan{A}, b = \tan{B}, c = \tan{C}$

Then the problem reduces to maximising

$\cos(2A) + \cos(2B) + 2\cos C$

subject to the constraint

$\cos(A+B+C) = 0$

and A,B,C are acute angles.

I don't know if this has a nice geometric interpretation, but if anyone has one, feel free to share it.

Edit: oh this was the solution described two posts above

inequality (1 Viewer)

s97127

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