RMO Algebra Prep | Srikanth B. Pai

(BMO 2005) Find the minimum possible value of $x^2 + y^2$ given that x and y are real numbers with $x \neq 0$, satisfying \[xy( x^2 − y^2 ) = x^2 + y^2.\]

Solution: Put $x=r\cos\theta$, $y=r\sin\theta$. We are supposed to minimize $r^2$ and the constraint reduces to \[r^2 \sin 4\theta = 4.\] So $r^2$ is minimized when $\sin 4\theta = 1.$ Thus the minimum value of $r^2$ is $4$.

[BMO 2008] Find all real values of $x, y$ and $z$ such that
\[(x+1)yz = 12\] \[(y+1)zx=4\] \[(z+1)xy = 4\]

Solution(Mrinal). We begin by subtracting (2) from (3), which yields \[x(y-z)=0.\] Clearly $x \neq 0$ as it would not satisfy the last two equations, and thus $y=z$. We proceed by subtracting (2) from (1), and adding (2) and (3), yielding \[y(y-x)=2xy(y+1)=8.\] Thus $x = \frac{4}{y(y+1)}$. Substituting this into the above, we get

\[\\ \begin{aligned} y\left(y-\dfrac{4}{y(y+1)}\right)&=8. \\ y^3+y^2-8y-12&=0, \\ (y+2)^2(y-3)&=0. \end{aligned} \\\]

Thus, the required solution sets are $(x, y, z) = (2, -2, -2) \ \text{and} \ \left(\frac{1}{3}, 3, 3\right).$

(BMO 1993) Let x,y,z be positive real numbers satisfying $\frac{1}{3} \leq xy+yz+zx \leq 3$ . Determine the range of values for xyz and x+y+z.

Proof(Rishon,Abhhi,Srikanth): We claim the range of $xyz$ is $(0,1]$. Let $\alpha > 0$ be a real. We choose $x=\alpha,y=z=\frac1{\alpha}$ then we see that $xy+yx+xz = 2+\frac1{\alpha^2}$ which satisfies the given inequality for $\alpha \geq 1$. Now \[xyz = \frac1{\alpha}.\] Note that for any real $r \in (0,1]$, $\alpha = \frac1{r}$ will give $xyz=r$. So we are done.

We claim the range of $x+y+z$ is $[1,\infty)$. By applying AM-GM to $x^2+y^2,y^2+z^2,x^2+z^2$ we see that \[x^2+y^2 + z^2 \geq xy + yz + xz.\]

This is equivalent to $(x+y+z)^2 \geq 3(xy+yz+xz) \geq 1$ . Since $x,y,z > 0$ we conclude $x+y+z \geq 1.$ Equality can be achieved for $x=y=z=\frac13,$ which is also satisfied by the given inequality (in the problem statement).

We will show that all the values above 1 will be achieved using intermediate value theorem (which can be found at the end of this solution).

If we choose $x=\frac13 + r,y=\frac13,z=\frac13$, then we see that \[xy+yx+xz = \frac{2r+1}3 .\] The given inequality reduces to $0\leq r \leq 4.$ Thus we see that the range of $x+y+z = 1+r$ is $[1,5]$ for $0\leq r \leq 4.$

Let $\alpha > 0$ be a real. Next we take $x=\alpha,y=z=\frac1{\alpha}$ then we see that $xy+yx+xz = 2+\frac1{\alpha^2}$ which satisfies the given inequality for $\alpha \geq 1$. Now \[x+y+z = \alpha + \frac{2}{\alpha} \geq 2\sqrt{2}.\] The last inequality is by AM-GM and equality is achieved at $\alpha = \sqrt{2}$. Now since $f(\alpha) = \alpha + \frac{2}{\alpha}$ is a continuous unbounded function, by IMVT its range is $[2\sqrt{2},\infty)$ .

Thus we conclude $[1,5] \cup [2\sqrt{2},\infty)=[1,\infty)$ is the range of $x+y+z$.

Appendix to Problem 1:

(Intermediate Value Theorem/IMVT) Let $f:[a,b] \to \mathbb{R}$ be a continuous function. If $\lambda$ is between $f(a), f(b)$ then there is an $x \in (a,b)$ such that $f(x) = \lambda$.

We shall also use the fact that sum, difference, product and quotient (as long as denominator does not vanish) of continuous functions are continuous. Additionally, polynomials are continuous functions.

(BMO 2004) Let S be a set of rational numbers with the following properties:

$\frac12 \in S$

If x ∈ S , then both $\frac1{x+1} \in S$ and $\frac{x}{x+1} \in S$.

Prove that S contains all rational numbers in the interval 0<x<1 .

Proof(Rishon): The proof is by strong induction on the denominator of a rational number between 0 and 1. The base case is $n=2$ and the only fraction with denominator 2 is $\frac12$. It is given that S contains this fraction.
Suppose we have shown that all rational numbers, in their simplest form with denominator strictly less than $n$, are in $S$. We break the calculation into three cases:

Case 1: If $k < n/2$, then $k < n-k$. Now \[\dfrac{k}{n-k}\in S \implies \dfrac{k}{n}=\dfrac{\frac{k}{n-k}}{\frac{k}{n-k}+1} \in S\] Case 2: If $k > n/2$, then $k > n-k$. Again we see \[\dfrac{n-k}{k}\in S \implies \dfrac{k}{n}=\dfrac{1}{\frac{n-k}{k}+1} \in S.\] Case 3: If $k = n/2$, then \[\dfrac{k}{n} = \dfrac12 \] which is in $S$ by the hypothesis of the problem. So we are done.

(BMO R2 2019) Find all functions $f$ from the positive real numbers to the positive real numbers for which $f( x ) \leq f( y )$ whenever $x \leq y$ and \[f( x^4 ) +f( x^2 ) +f( x ) +f( 1 ) = x^4 + x^2 +x+1.\] for all $x>0$.

Proof (Rishon, Lauren, Abhhi and Srikanth): Clearly $f(1)=1.$ If we define $g(x) = f(x) - x$, then we see $g(x^4)+g(x^2)+g(x)=0.$ But $x=y,x=y^2$ and subtract to get \[g(y^8) = g(y)\] for all real $y > 0$. Note that this means \[\displaystyle g(x^{8^n}) =g(x)=g(x^{\frac1{8^n}})\] for all $n$ and for all $x>0$.

Now we show that $g\equiv0$, by analysing two cases.

Case 1: For $x < 1$, since $g(1)=0$, and the non-decreasing nature of $f$ we get: \[g(x) + x \leq g(1)+1 = 1\] Now we note that $x^{\frac1{8^n}}<1$ since $x<1$. So \[g(x) + x^{\frac1{8^n}} = g(x^{\frac1{8^n}}) + x^{\frac1{8^n}} \leq 1.\] Thus $g(x) \leq 1 - x^{\frac1{8^n}}$ for all $n$. By applying $lim_{n \to \infty}$ on both sides and noting $\lim_{n\to \infty} x^{\frac1{8^n}} = 1$ for $x>0$, we get \[g(x) \leq 0.\]If $x<1$, then $x^2,x^4 < 1$. So $g(x^4),g(x^2),g(x)\leq 0$. But $g(x^4)+g(x^2)+g(x)=0$ and thus each term is individually zero. So we conclude $g(x) = 0$ for $0 < x < 1$.

Case 2: For $x > 1$, since $g(1)=0$, \[g(x) + x \geq 1\] Now we note that $x^{\frac1{8^n}}>1$ since $x>1$. So \[g(x) + x^{\frac1{8^n}} = g(x^{\frac1{8^n}}) + x^{\frac1{8^n}} \geq 1.\] Thus $g(x) \geq 1 - x^{\frac1{8^n}}$ for all $n$. By applying $\displaystyle \lim_{n \to \infty}$ on both sides and noting $\displaystyle \lim_{n\to \infty} x^{\frac1{8^n}} = 1$ for $x>0$, we get \[g(x) \geq 0.\] If $x>1$, then $x^2,x^4 > 1$. So $g(x^4),g(x^2),g(x)\geq 0$. But $g(x^4)+g(x^2)+g(x)=0$ and thus each is individually zero. So we conclude $g(x) = 0$ for $x >1$.

Thus we have showed that $g\equiv0$.

(RMO 2017) Let $x, y, z$ be real numbers, each greater than 1. Prove that \[\dfrac{x+1}{y+1} + \dfrac{y+1}{z+1} + \dfrac{z+1}{x+1} \leq \dfrac{x-1}{y-1} + \dfrac{y-1}{z-1} + \dfrac{z-1}{x-1}.\]

Proof(Mrinal). Let us assume that $x \geq y \geq z$. We are to show that \[\sum_{cyc} \left(\dfrac{x-1}{y-1}-\dfrac{x+1}{y+1}\right) = \sum_{cyc} \left( \dfrac{x-y}{y^2-1} \right) \geq 0.\] Begin by calling the latter quantity $G$. Since $x \geq y \geq z$, we have \[G=\dfrac{x-y}{y^2-1}+\dfrac{y-z}{z^2-1}+\dfrac{z-y+y-x}{x^2-1}.\] This is equivalent to \[G=(y-z)\left( \dfrac{1}{z^2-1}-\dfrac{1}{x^2-1}\right)+(x-y)\left( \dfrac{1}{y^2-1}-\dfrac{1}{x^2-1}\right).\] Since $x \geq y, y \geq z$ and $x \geq z$, each expression bounded by paranthesis is non-negative. So we conclude that $G \geq 0$, and we are done. A similar proof is needed for $x \geq z \geq y$ which we leave to the reader.

(RMO 2017) Let $P(x)=x^2+\frac{1}{2}x+b$ and $Q(x)=x^2+cx+d$ be two polynomials with real coefficients such that $P(x)Q(x)=Q(P(x))$ for all real $x$. Find all the real roots of $P(Q(x))=0.$

Claim: The only real roots are $-1$ and $\frac12$.

Proof (Mrinal). The given condition is equivalent to \[\left(x^2+\dfrac{x}{2}+b\right)\left(x^2+cx+d \right)=\left( x^2+\dfrac{x}{2} +b\right)^2+c\left( x^2+\dfrac{x}{2}+b\right)+d.\] We now proceed by comparing coefficients of $x^3, x^2$ and constants of both sides (comparing coeffient of $x$ is redundant).

Comparing the coefficient of $x^3$ gives \[c+\dfrac{1}{2}=1 \implies c=\dfrac{1}{2}.\]
Comparing the coefficient of $x^2$ gives \[d+\dfrac{c}{2}+b=\dfrac{1}{4}+2b+c.\] On substituting $c=\frac12$, we get \[d=\dfrac{1}{2}+b.\]
Comparing constants, we get \[bd=b^2+\dfrac{b}{2}+d.\] Substituting $d=\frac12 + b$, we get $d=0, b=-\frac{1}{2}.$

We now set $P(Q(x))=0$ as follows.

\[\\ \begin{aligned} \left( x^2+\dfrac{x}{2} \right)^2+\dfrac{1}{2}\left( x^2+\dfrac{x}{2} \right)-\dfrac{1}{2}&=0 \\ 4x^4+4x^3+3x^2+x-2&=0. \end{aligned} \\\]

On inspection, we see that $(x+1)$ is a factor of the above. Performing long division yields \[4x^4+4x^3+3x^2+x-2=(x+1)(4x^3+3x-2).\] Again, by inspection, we see that $(2x-1)$ is a factor, giving us the final factorisation of \[4x^4+4x^3+3x^2+x-2=(x+1)(2x-1)(2x^2+x+2).\] Since the quadratic has no real roots, we conclude that the only real roots are $-1$ and $\frac{1}{2}$.

Appendix to Problem 1:

Enjoy Reading This Article?