# Andy Sandbox

Given \(a,\) \(b,\) and \(c\) are real numbers such that \(a^6+b^6+c^6=64,\) what is the maximum value of \(4a^2c+12ab^2+3bc^2\ ?\)

The expressions can be related through Hölder's inequality:

\[\begin{align} (4^2+12^2+3^2)^\frac{1}{2} (a^6+b^6+c^6)^\frac{1}{3} (c^6+a^6+b^6)^\frac{1}{6} &\ge 4a^2c+12ab^2+3bc^2 \\ (169)^\frac{1}{2} (64)^\frac{1}{3} (64)^\frac{1}{6} &\ge 4a^2c+12ab^2+3bc^2 \\ 104 &\ge 4a^2c+12ab^2+3bc^2. \end{align}\]

Thus, the maximum value of the expression is \(\boxed{104}.\)

Given \(\triangle ABC,\) place point \(U\) on \(\overline{BC}\) such that \(\overline{AU}\) bisects \(\angle A,\) and place point \(V\) on \(\overline{AC}\) such that \(\overline{BV}\) bisects \(\angle B.\) Let \(I\) be their point of intersection. Then place point \(X\) on \(\overline{BC}\) such that \(\overline{IX} \perp \overline{BC},\) place point \(Y\) on \(\overline{AC}\) such that \(\overline{IX} \perp \overline{AC},\) and place point \(Z\) on \(\overline{AB}\) such that \(\overline{IZ} \perp \overline{AB}.\) Finally, place point \(W\) on \(\overline{AB}\) such that \(\overline{CW}\) passes through point \(I.\)

\(\triangle AIY\) and \(\triangle AIZ\) have the following congruences:

- \(\angle AYI \cong \angle AZI\) because they are both right angles.
- \(\angle IAY \cong \angle IAZ\) because \(\overline{AI}\) is the angle bisector.
- \(\overline{AI} \cong \overline{AI}\) because of the reflexive property of congruence.
Thus, by AAS, \(\triangle AIY \cong \triangle AIZ.\) In a similar fashion, it can be proven that \(\triangle BIX \cong \triangle BIZ.\) Then, by CPCTC (congruent parts of congruent triangles are congruent) and the transitive property of congruence, \(\overline{IX} \cong \overline{IY} \cong \overline{IZ}.\)

Now \(\triangle CIX\) and \(\triangle CIY\) have the following congruences:

- \(\overline{IX} \cong \overline{IY},\) as stated earlier.
- \(\angle CXI \cong \angle CYI\) because they are both right angles.
- \(\overline{CI} \cong \overline{CI}\) because of the reflexive property of congruence.
Thus, by HL, \(\triangle CIX \cong \triangle CIY.\) By CPCTC, \(\angle ICX \cong \angle ICY.\) Hence, \(\overline{CW}\) is the angle bisector of \(\angle C,\) and all three angle bisectors meet at point \(I.\)

Since \(\overline{IX} \cong \overline{IY} \cong \overline{IZ},\) there exists a circle centered at \(I\) that passes through \(X,\) \(Y,\) and \(Z.\) Furthermore, since these segments are perpendicular to the sides of the triangle, the circle is internally tangent to the triangle at each of these points. Hence, the incentre is located at point \(I.\)