# Condition ax+by+cz≤xyz, expression αx+βy+γz

Haha, so I am back after a long time, it feels so nice to write something again on my good old blog. I was kind of busy with school exams and different mock exams for IIT-JEE in my coaching institute (NOT FIITJEE) and also a maths talent exam (which had a total shi**y question paper mostly full of calculations). So after all this business, I am back to maths. It’s not as if I was slacking off too much, but I really got into playing osu!

Today is the first day of Durga Puja, known as Mahalaya; and therefore a holiday. The month of November of a 12th grader in India is the busiest time of the year, full of different exams – competitive and school. So maybe it will be another long wait for this blog before it gets another update.

kk, I guess that concludes my set of excuses for not updating the blog for a long time. However, the title and the topic is totally random. Obviously it’s on inequalities(because it’s random), and it is on a method for solving problems of the type $ax+by+cz\leq xyz\implies \alpha x+\beta y+\gamma z\geq k$ where $x,y,z$ are positive real numbers and $a,b,c,\alpha,\beta, \gamma$ are given constants.

Introduction.

I don’t consider this necessary, but probably I should show one inequality of this kind. This is an excerpt from my “Topic of Tran Quoc Anh” file, which is under development. This inequality probably inspired me to find out the technique of which I am talking now.

Hmm, now you may ask some questions, like how the hell did I come up with the exact coefficients and why did I use $10=9+1$ and not $10=1+9.$ Well, when I was trying to solve it in a way other than how the author came up with the problem, I was trying different ways of reducing the variables. The strategy that I used was to boil down the inequality on 3 variables to 1 variable. The same procedure will work for our general case.

Variable Elimination.

On a totally general note, this is quite useful in non-symmetric inequalities. In fact, this is one definite method which should work always. We try to reduce a $n$ variable inequality to lesser number of variables. One method that people use quite frequently is the Entirely Mixing Variables Method(EMV), and its stronger version: The Stronger Mixing Variables Method(SMV). Now, this one is quite unrelated to MV and is quite a raw form of elimination, but still quite useful.

The General Problem.

Well, it’s obvious that not all $a,b,c,\alpha, \beta,\gamma$ will abide by a minimum value. We consider, for the sake of simplicity that $x,y,z,a,b,c,\alpha,\beta,\gamma>0.$ So, the problem that we will look into is:

Let $x,y,z$ be positive real numbers such that for some constants $a,b,c>0$ we have $ax+by+cz\leq xyz.$ Then find the conditions for which the expression $P(x,y,z)\equiv \alpha x+\beta y+\gamma z$ has a minimum and also determine the minimum.

The “General” Solution.

Again, we follow the same method as what we used for the example problem.
Note that $z\geq \dfrac{ax+by}{xy-c}.$ Then,

\displaystyle \begin{aligned}P\equiv \alpha x+\beta y+\gamma z&\geq \alpha x+\beta y+\gamma \cdot \frac{ax+by}{xy-c}\\&=\alpha x+\frac{\beta}{x} \cdot xy+\frac{\gamma}{x}\cdot \frac{ax^2+bxy}{xy-c}\\&=\alpha x+ \frac{\beta c+\gamma b}{x}+\frac 1x\left[\beta(xy-c)+\gamma \cdot \frac{ax^2+bc}{xy-c}\right];\end{aligned}

And now we know how to eliminate $y$ from our expression. What we get after applying AM-GM is that

\displaystyle \begin{aligned}P&\geq \alpha x+\frac{\beta c+\gamma b}{x}+2\sqrt{\beta\gamma\left(\frac{ax^2+bc}{x^2}\right)}\\&=\alpha x+\frac{\beta c+\gamma b}{x}+2\sqrt{\beta\gamma\left(a+\frac{bc}{x^2}\right)}.\end{aligned}

Hmm, this is a one-variable inequality. We have been able to generalise our method till this line, but after this I was just lucky to have stumbled upon $10$ inside the square root, but right now there is a $\beta \gamma.$ Also, for the equality to hold till this point, we have made a hell lot of assumptions. In each step that we write a $\geq,$ we need to verify for the equality case. So for people who are too lazy to push this thing into an elegant solution, might as well differentiate the expression. But, the AM-GM solution is way more spectacular than a calculus bash; also Cauchy-Schwarz and AM-GM are my favourite inequalities – so anyway let’s continue.

So, we want to find $k,l$ such that $ak^2+bcl^2=4\beta \gamma.$ Assume that we had been able to find some suitable $k,l$ (I am not explaining what “suitable” means right now). Then we can get

\displaystyle \begin{aligned}P&\geq \alpha x+\frac{\beta c+\gamma b}{x}+\sqrt{(ak^2+bcl^2)\left(a+\frac{bc}{x^2}\right)}\\&\geq \alpha x+\frac{\beta c+\gamma b}{x}+ak+\frac{bcl}{x}\\&=ak+\alpha x+\frac{\beta c+\gamma b+ bcl}{x}\\&\geq ak+2\sqrt{\alpha\left(\beta c+\gamma b+bcl\right)}.\end{aligned}

So one thing is for sure. If we can find $k,l$ and if the $a,b,c,\alpha,\beta,\gamma$ all abide by our restrictions, then this must be the required minimum that we are seeking. So now we have to find the constraints.

Constraints.

Start with the first line of the solution. I suddenly wrote $z\geq \dfrac{ax+by}{xy-c},$ but this will be true only if $xy>c.$ However, $xyz\geq ax+by+cz>cz$ is obvious. But this will help us determine $z$ from $x$ and $y,$ so let’s call this constraint the $z$– determination constraint.
Now come to the second $\geq$ of the solution. For equality to hold, we must have $(xy-c)^2=\dfrac{\gamma}{\beta}\cdot (ax^2+bc).$ Let’s call this the $y$– determination constraint.
Now the third prey. For equality in the Cauchy-Schwarz step; ie $\displaystyle (ak^2+bcl^2)\left(a+\frac{bc}{x^2}\right)\geq \left(ak+\frac{bcl}{x}\right)^2,$ we can verify that the condition $x$ must follow is $x=\dfrac{k}{l}.$
Fourth condition. In the final step, for equality to hold we should have $x=\sqrt{\dfrac{\beta c+\gamma b+bcl}{\alpha}}.$

The last two constraints will let us solve for $k,l.$ What we get is that
$\dfrac{k^2}{l^2}=\dfrac{\beta c+\gamma b +bc l}{\alpha};$
Which simplifies to the following cubic in $l;$ which I would like to call the Characteristic cubic of the problem:

$\displaystyle l^3+\left(\frac{\alpha}{a}+\frac{\beta}{b}+\frac{\gamma}{c}\right)l^2-\frac{4\alpha\beta\gamma}{abc}=0.$

Conditions for Minima.

So we have obtained some necessary conditions for the minimum to exist. For example, the real root $l_1$ of the characteristic equation must be positive; and also the $l_1$ that we obtain this way must give a corresponding $k,$ which is verified when $\dfrac{4\beta\gamma}{bc}\geq l^2.$ But, note that if $l_1$ is a positive root of the characteristic cubic then

$\displaystyle l_1^2\cdot \frac{\alpha}{a}-\frac{4\alpha\beta\gamma}{abc}< l_1^3+l_1^2\sum_{cyc}\frac{\alpha}{a}-\frac{4\alpha\beta\gamma}{abc}=0;$ so that any positive root of this cubic will always lead us to a corresponding value of $k,$ thus fixing the $x,y,z$ for which the minimum is attained. So finally what we have to check for is the existence of a positive real root of the characteristic cubic.

Take $r=\frac{\alpha}{a},$ $s=\frac{\beta}{b},$ $t=\frac{\gamma}{c}.$ Then the cubic is $f(x)=x^3+(r+s+t)x^2-4rst.$ Now $f'(x)=3x^2+2x(r+s+t)$ so $f$ has two real points of inflection : $x=0$ and $x=-\frac{2}{3}(r+s+t).$ Now $f(0)=-4rst<0$ and $f$ is increasing for $x>0,$ so this cubic equation will always have one positive root. It may have two or no negative roots. So any $a,b,c,\alpha,\beta, \gamma>0$ satisfy and permit such a minimum value.

The value of $l$ which satisfies the cubic equation can be found out to be a huge expression.

So we can take any $a,b,c,\alpha,\beta,\gamma$ we like and then make a new inequality using the characteristic cubic. $\Box$

This was not too tl;dr, but the method is awesome, so I wrote down this thing in an hour. The next post should be on some other more interesting topic. 🙂