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Problem A. 889. (October 2024)

A. 889. Let \(\displaystyle W, A, B\) be fixed real numbers with \(\displaystyle W>0\). Prove that the following statements are equivalent.

Proposed by Ákos Somogyi, London

(7 pont)

Deadline expired on November 11, 2024.

For ease of notation, define the viable set of \(\displaystyle (x,y,z)\) triples for a given \(\displaystyle W\) as the following:

\(\displaystyle H_{W}:=\left\{ \left(x,y,z\right):x,y,z\geq0\text{ and }x+y\leq z+W,\ x+z\leq y+W,\ y+z\leq x+W\right\} \)

Trivial bounds for the individual variables:

\(\displaystyle \left(x+y\right)+\left(x+z\right)\leq\left(z+W\right)+\left(y+W\right)\Rightarrow x\leq W,\)

and by symmetry we obtain \(\displaystyle y,z\le W\).

\(\displaystyle \left(x,x,W\right)\in H_{W}\) for \(\displaystyle x\in\left[0,W\right]\), the inequality for this selection of variables reads:

\(\displaystyle AWx^{2}+B\geq2x^{2}+W^{2}\qquad\begin{cases} x=0\Rightarrow & \quad B\geq W^{2}\\ x=W\Rightarrow & \quad AW^{3}+B\geq3W^{2} \end{cases} \)

This shows that the first statement implies the second. It remains to prove that the converse is also true.

Observe the following:

\(\displaystyle Axyz+B=\left(\frac{2xyz}{W}+W^2\right)+\frac{AW^3+B-3W^2}{W^3}xyz+\frac{B-W^2}{W^3}(W^3-xyz). \)

The second and third terms are non-negative on the right hand side, since \(\displaystyle 0\le x,y,z\le W\) implies \(\displaystyle 0\le xyz\le W^3\).

Hence it is enough to prove that \(\displaystyle \frac{2xyz}{W}+W^{2} \geq x^2+y^2+z^2\). First divide both sides by \(\displaystyle W^{2}\) and apply the substitution \(\displaystyle a=\frac{x}{W},b=\frac{y}{W},c=\frac{z}{W}\) – these are still non-negative as \(\displaystyle W>0\).

\(\displaystyle \frac{2xyz}{W}+W^{2}\geq\left(x^{2}+y^{2}+z^{2}\right)\Rightarrow2\frac{x}{W}\cdot\frac{y}{W}\cdot\frac{z}{W}+1\geq\left(\frac{x}{W}\right)^{2}+\left(\frac{y}{W}\right)^{2}+\left(\frac{z}{W}\right)^{2}. \)

Conditions are changed to

\(\displaystyle a+b\leq c+1,\quad a+c\leq b+1,\quad b+c\leq a+1 \)

(i.e. we're working with \(\displaystyle W=1\) so we need to prove for \(\displaystyle \left(a,b,c\right)\in H_{1}\)), individual bounds to \(\displaystyle 0\leq a,b,c\leq1\) and we need to prove

\(\displaystyle \left(2abc+1\right)\geq a^{2}+b^{2}+c^{2}\Leftrightarrow \left(2abc+1\right)-\left(a^{2}+b^{2}+c^{2}\right)\geq0.\)

Rearrange:

\(\displaystyle \left(2abc+1\right)-\left(a^{2}+b^{2}+c^{2}\right)=\left(1-b^{2}\right)\left(1-c^{2}\right)-\left(a-bc\right)^{2}\geq0. \)

Without loss of generality (due to symmetry) we can assume \(\displaystyle 1\geq c\geq b\geq a\geq\max\left(0,b+c-1\right) \). The LHS is a quadratic (continuous) function in \(\displaystyle a\) and has a maximum at \(\displaystyle a=bc\) and strictly increasing on \(\displaystyle \left(-\infty,bc\right]\) and strictly decreasing on \(\displaystyle \left[bc,\infty\right)\). Hence on any closed, finite interval \(\displaystyle \left[u,v\right]\) the minimum of the LHS is attained at one of the endpoints of the interval. To verify the inequality, enough to check the endpoints of the interval for viable \(\displaystyle a\)s for any fixed \(\displaystyle b\) and \(\displaystyle c\).

Upper bound, i.e. \(\displaystyle a=b\), then \(\displaystyle 1\geq c\geq b\geq b^2\) and thus \(\displaystyle 1+c\geq 2b^2\)

\(\displaystyle \left(1-b^{2}\right)\left(1-c^{2}\right)-\left(a-bc\right)^{2}=\left(1-b^{2}\right)\left(1-c^{2}\right)-\left(b-bc\right)^{2}=(1-c)\left(c+1-2b^{2}\right)\geq0 \)

Equality holds if \(\displaystyle a=b=c=1\) in this case.

Lower bound case 1: \(\displaystyle a=0\Rightarrow\) \(\displaystyle 0\geq b+c-1\), i.e \(\displaystyle 1\geq b+c\):

\(\displaystyle \left(1-b^{2}\right)\left(1-c^{2}\right)-\left(a-bc\right)^{2}=\left(1-b^{2}\right)\left(1-c^{2}\right)-\left(0-bc\right)^{2}=1-b^{2}-c^{2}\geq1-b-c\geq0 \)

Equality holds if \(\displaystyle 1=b^2+c^2\) and \(\displaystyle b+c\leq1\) i.e. \(\displaystyle a=b=0\) and \(\displaystyle c=1\).

Lower bound case 2: \(\displaystyle a=b+c-1\Rightarrow\) \(\displaystyle 0\leq b+c-1,i.e1\leq b+c:\)

\(\displaystyle \left(1-b^{2}\right)\left(1-c^{2}\right)-\left(a-bc\right)^{2}=\left(1-b^{2}\right)\left(1-c^{2}\right)-\left(b+c-1-bc\right)^{2}=2(1-b)(1-c)(b+c)\geq0 \)

Equality holds if \(\displaystyle c=1\), \(\displaystyle a=b+c-1=b\).

This shows that the function \(\displaystyle \left(1-b^{2}\right)\left(1-c^{2}\right)-\left(a-bc\right)^{2}\) is non-negative at the endpoints of the interval valid for \(\displaystyle a\). Equality holds with \(\displaystyle b+c=1\) and \(\displaystyle a=0\) This implies for any \(\displaystyle \left(a,b,c\right)\in H_{1}\), \(\displaystyle 2abc+1\geq a^{2}+b^{2}+c^{2}\) holds.

Therefore, for any \(\displaystyle \left(A,B\right)\) with \(\displaystyle B\geq W^{2}\), \(\displaystyle AW^{3}+B\geq3W^{2}\) the inequality holds for all \(\displaystyle \left(x,y,z\right)\in H_{W}\).

Note 1: Our rearrangement of \(\displaystyle Axyz+B\) needs some explanation. First consider the equality case:

\(\displaystyle B=W^{2}\qquad AW^{3}+B=3W^{2}\Rightarrow A=\frac{2}{W} \)

We want to keep the inequalities for \(\displaystyle A,B,W\) true while searching for a decomposition with few non-negative terms and a remaining inequality (which we think could be the boundary case). We try to preserve \(\displaystyle AW^{3}+B\geq3W^{2}\). So we need some \(\displaystyle \delta,\gamma\) reals such that \(\displaystyle AW^{3}+B=\left(A+\delta\right)W^{3}+B-\gamma\) holds and

\(\displaystyle Axyz+B\geq\left(A+\delta\right)xyz+\left(B-\gamma\right)\Rightarrow\frac{\gamma}{\delta}\geq xyz \)

\(\displaystyle Axyz+B-\left(A+\delta\right)xyz-\left(B-\gamma\right)=\gamma-\delta xyz \)

\(\displaystyle AW^{3}+B=\left(A+\delta\right)W^{3}+B-\gamma\Rightarrow\gamma=\delta W^{3}\Rightarrow\frac{\gamma}{\delta}=W^{3} \)

Hence by \(\displaystyle (i)\) \(\displaystyle \frac{\gamma}{\delta}=W^{3}\geq xyz\) and \(\displaystyle \delta\geq0\) we can write:

\(\displaystyle AW^{3}+B=\delta\left(W^{3}-xyz\right)+\left(A+\delta\right)xyz+\left(B-\delta W^{3}\right) \)

Since the second condition is \(\displaystyle B\geq W^{2}\), so by choosing \(\displaystyle B-\delta W^{3}=W^{2}\Rightarrow\delta=\left(\frac{B-W^{2}}{W^{3}}\right)\geq0\) due to \(\displaystyle B\geq W^{2}\) makes \(\displaystyle \delta\left(W^{3}-xyz\right)\) non-negative. Furthermore \(\displaystyle Axyz+B\geq\left(A-\varepsilon\right)xyz+B\), so we can try to factor out the equality case for the bounds on \(\displaystyle A,B,W\)

\(\displaystyle AW^{3}+B=\left(\frac{B-W^{2}}{W^{3}}\right)\left(W^{3}-xyz\right)+\left(A+\left(\frac{B-W^{2}}{W^{3}}\right)\right)xyz+W^{2}= \)

\(\displaystyle =\left(\frac{B-W^{2}}{W^{3}}\right)\left(W^{3}-xyz\right)+\left(A+\left(\frac{B-W^{2}}{W^{3}}\right)-\frac{2}{W}\right)xyz+\frac{2}{W}xyz+W^{2}= \)

\(\displaystyle =\left(\frac{B-W^{2}}{W^{3}}\right)\left(W^{3}-xyz\right)+\left(\frac{AW^{3}+B-3W^{2}}{W^{3}}\right)xyz+\frac{2}{W}xyz+W^{2} \)

Note 2: We reduced the problem to A.371. https://www.komal.hu/feladat?a=feladat&f=A371&l=hu

Statistics:

15 students sent a solution.
7 points:Bodor Mátyás, Forrai Boldizsár, Gyenes Károly, Holló Martin, Keresztély Zsófia, Kocsis 827 Péter, Morvai Várkony Albert, Szakács Ábel, Tianyue DAI, Varga Boldizsár, Vigh 279 Zalán.
4 points:1 student.
3 points:2 students.
2 points:1 student.

Problems in Mathematics of KöMaL, October 2024