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### A beautiful generalization of the Nesbitt Inequality

I want to discuss a beautiful inequality, that is a generalization of the famous Nesbitt inequality:

(Romanian TST) For positive $a,b,x,y,$, prove that $\frac{x}{ay+bz}+\frac{y}{az+bx}+\frac{z}{ax+by}\geq \frac{3}{a+b}$

Clearly, if $a=b$, then we get Nesbitt’s inequality, which states that

$\frac{x}{y+z}+\frac{y}{z+x}+\frac{z}{x+y}\geq \frac{3}{2}$.

This is question 14 on Mildorf’s “Olympiad Inequalities”, and its solution comprises finding a factor to multiply this expression with, almost out of thin air, and then use Cauchy Schwarz and AM-GM inequalities to prove the assertion. My solution is the following:

On interchanging $a$ and $b$, the right hand side remains the same. However, the left hand side becomes

$\frac{x}{az+by}+\frac{y}{ax+bz}+\frac{z}{ay+bx}\geq \frac{3}{a+b}$

On adding these two inequalities, we get

$\frac{x}{ay+bz}+\frac{x}{az+by}+\frac{y}{az+bx}+\frac{y}{ax+bz}+\frac{z}{ax+by}+\frac{z}{ay+bx}\geq \frac{6}{a+b}$

Multiplying both sides by $\frac{1}{2}(a+b)$ and then adding $6$ on both sides, we get

$2(x+y+z)(\frac{1}{x+y}+\frac{1}{y+z}+\frac{1}{z+x})\geq 9$

This is obviously true by Cauchy Schwarz. We will explain below how we got this expression.

Let us see what happens to $\frac{x}{ay+bz}+\frac{x}{az+by}$ in some detail. After multiplying by $\frac{1}{2}(a+b)$ and adding $2$, we get

$\frac{\frac{1}{2}(a+b)x+ay+bz}{ay+bz}+\frac{\frac{1}{2}(a+b)x+az+by}{az+by}\geq 2\frac{(a+b)(x+y+z)}{(a+b)(y+z)}=2\frac{(x+y+z)}{y+z}$.

EDIT: I assumed that this was obviously true. However, it is slightly non-trivial that this is true. For $\frac{a}{b}+\frac{c}{d}\geq 2\frac{a+c}{b+d}$, the condition that should be true is that $(b-d)(bc-ad)\geq 0$. This is true in our case above.

After adding the other terms also, we get

$2(x+y+z)(\frac{1}{x+y}+\frac{1}{y+z}+\frac{1}{z+x})$

As pointed above, this is clearly $\geq 9$ by Cauchy-Schwarz.

Hence proved

Note: For the sticklers saying this isn’t a rigorous proof, a rigorous proof would entail us assuming that

$\frac{x}{ay+bz}+\frac{y}{az+bx}+\frac{z}{ax+by}< \frac{3}{a+b}$, and then deriving a contradiction by proving that

$2(x+y+z)(\frac{1}{x+y}+\frac{1}{y+z}+\frac{1}{z+x})< 9$, which is obviously false

### A small note on re-defining variables to prove inequalities

I just want to record my solution to the following problem, as it is different from the one given online.

For $a,b,c,d$ positive real numbers, prove that $\frac{1}{a}+\frac{1}{b}+\frac{4}{c}+\frac{16}{d}\geq \frac{64}{a+b+c+d}$

This has a fairly straight forward solution using Cauchy-Schwarz inequality, which for some reason I did not think of.

The way that I solved it is that I re-defined the variables: let $a=8a', b=8b', c=16c'$ and $d=32 d'$. Then this is equivalent to proving that $\frac{1}{8}\frac{1}{a'}+\frac{1}{8}\frac{1}{b'}+\frac{1}{4}\frac{1}{c'}+\frac{1}{2}\frac{1}{d'}\geq \frac{1}{\frac{a'}{8}+\frac{b'}{8}+\frac{c'}{4}+\frac{d'}{2}}$

This is easily seen to be a consequence of Jensen’s inequality, as $\frac{1}{x}$ is a convex function for positive $x$.

### Some solutions

I solved two math problems today. The solutions to both were uniquely disappointing.

The first problem was the first problem from IMO 1986:

Let $d$ be a number that is not $2,5$ or $13$. Prove that out of the set $\{2,5,13,d\}$, we can select two different numbers $a,b$ such that $ab-1$ is not a square.

A quick check would show that $2.5-1, 2.13-2$ and $5.13-1$ are all squares. This shows that of the two numbers that we select from $\{2,5,13,d\}$, one of them has to be $d$. We essentially need to prove that for $d\neq 2,5,13$, $2d-1, 5d-1$ and $13d-1$ are not all squares (a maximum of two of them might be. But all three cannot).

I tried figuring out some of the implications of all of them being squares, but couldn’t come up with anything concrete, as I didn’t write anything at all, and just tried to devise a strategy mentally. Ultimately, I came up with possibly the most mundane proof of all time.

Proof: Consider squares $\mod 16$ (I couldn’t come up with any contradiction up until $\mod 8$). The possible residues for squares $\mod 16$ are $0,1,4,9$. Now my strategy was to plug in all possible residues $\mod 16$ in place of $d$, and checking that in every such instance, at least one of $2d-1, 5d-1$ and $16d-1$ gave a residue that was not $0,1,4,9\mod 16$. This I could successfully verify. Hence proved.

The solution that I could find in the IMO compendium was much smarter, but involved some algebra that would obviously involve putting pen to paper.

The second question that I solved today was proving the Koszul formula, which states that

$2g(\nabla_X Y,Z)=D_Xg(Y,Z)+D_Yg(X,Z)-D_Zg(X,Y)-g([Y,X],Z)-g([X,Z],Y)+g([Z,Y],X)$

I’d seen proofs before in textbooks, that would involve writing down expansions, and then combining all of them in some clever way. The way that I came up with was pretty algorithmic, and perhaps even more elementary. It is simply the following: Take $g(\nabla_X Y,Z)$. Use the torsion free condition to write it as $g(\nabla_Y X,Z)+g([X,Y],Z)$. Now use the formula for connection acting on a tensor to write $g(\nabla_Y X,Z)$ as $D_Y g(X,Z)-g(X,\nabla_Y Z)$. Performing the two operations, we have $D_Yg(X,Z)-g(X,\nabla_Y Z)+g([X,Y],Z)$. Performing this operation of first using the torsion free condition and then the formula for a connection on a tensor three times, we get the Koszul formula.

I’ve been obsessed with Math Olympiads and other kinds of competitive math for a loong time now. Although I should be doing more serious mathematics now that can actually get me a job, I shall ignore good sense and start my series of posts on Math Olympiad problems!

The first problem that I want to write about is a relatively easy one from IMO (International Math Olympiad) 1974. I think it was a long-listed problem, and didn’t actually make it to the test. It asks to prove the fact that

$\displaystyle 343|2^{147}-1$

(that $343$ divides $2^{147}-1$)

The important thing in this problem is to figure out that $343=7^3$ and $147=3.7^2$. Hence, we’re being asked to prove that $7^3|2^{3.7^2}-1$.

We have the following theorem for reference: If $a$ and $b$ are co-prime, then $a^{\phi(b)}\equiv 1\pmod b$, where $\phi(b)$ is the number of natural numbers that are co-prime with $b$. Fermat’s Little Theorem is a special case of this.

Now on to the calculations and number crunching. We have fleshed out the main idea that we’ll use to attack the problem. We can see that $\phi(7^3)=7^3-7^3/7=6.7^2$. Hence, as $2$ is co-prime with $7^3$, $2^{6.7^2}-1\equiv 0\pmod {7^3}$. We can factorize this as $(2^{3.7^2}+1)(2^{3.7^2}-1)\equiv 0\pmod {7^3}$. Now we have to prove that $7^3$ does not have any factors in common with $2^{3.7^2}+1$. As $2^3\equiv 1\pmod 7$, $2^{3.7^2}\equiv 1\pmod 7$ too. Hence, $2^{3.7^2}+1\equiv 2\pmod 7$. This shows that $2^{3.7^2}+1$ does not have any factors in common with $7^3$, and leads us to conclude that $7^3|(2^{3.7^2}-1)$, or $343|2^{147}-1$. Hence proved.

I realize that the formatting is pretty spotty. I shall try and re-edit this in the near future to make it more readable.