Cubes Constrained


Cubes Constrained


Let $a^3=x,b^3=y,\,$ then $x+y=2.\,$ Consider function $f(t)=t^{\frac{1}{3}},\,$ $t\ge 0.\,$ The function is concave such that

$\displaystyle f(x)-f(1)\le (x-1)f'(1),$

or, $\displaystyle x(f(x)-f(1))\le x\frac{x-1}{3}.\,$ Similarly,

$\displaystyle\begin{align}y(f(y)-f(1)\le y\frac{y-1}{3}\;\text{and}\\ xy(f(xy)-f(1)\le xy\frac{xy-1}{3}, \end{align}$

from which

$\displaystyle \begin{align} 3(a^4+b^4)+2a^4b^4&=3\left(x^{\frac{4}{3}}+y^{\frac{4}{3}}\right)+2x^{\frac{4}{3}}y^{\frac{4}{3}}\\ &=3(xf(x)+yf(y)]+2xyf(xy)\\ &\le 3(x+y)f(1)+x(x-1)+y(y-1)\\ &\qquad\qquad+2xyf(1)+2xy\frac{xy-1}{3}\\ &=6+x^2+y^2-2+2xy+2xy\frac{xy-1}{3}\\ &=4+(x+y)^2+2xy\frac{xy-1}{3}\\ &=8+2xy\frac{xy-1}{3}\le 8, \end{align}$

because, by the AM-GM inequality, $xy-1\le 0.$


Marian Dinca has kindly communicated to me his solution to an year-old post by Leo Giugiuc at the The School Yard Olympiad facebook group. Leo credits the problem to Andrei Eckstein.


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