For positive real numbers x, y satisfy the condition $ x + y \le $1. Find the maximum value of the expression $$ P = {{x} ^ {2}} - \frac {3} {4x} - \frac {x} {y} $$ (Please help not to use derivative because this way I solved. Only use basic inequalities)


closed as off-topic by Martin R, Arnaud D., Namaste, GNUSupporter 8964民主女神 地下教會, polfosol Apr 9 '18 at 13:01

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$$P(x,y)-P\left(\frac{1}{2},\frac{1}{2}\right)=-\frac{(1-x-y)x}{y(1-x)}-\frac{(2x-1)^2\left(x^2+3\right)}{4x(1-x)}\leq 0$$


Clearly, for $y_0<y_1$ we get

$$x^2- \frac {3} {4x} - \frac {x} {y_0}<x^2- \frac {3} {4x} - \frac {x} {y_1}$$

for any $x$, so this means the maximum will have the largest possible $y$. Since $x+y\leq 1$, this must be $y=1-x$. This means we are searching for the maximum of


where $0<x<1$. We will first rewrite it:


Now since $4x(1-x)$ is symmetric over $\tfrac12$, we will substitute $z=x-\tfrac12$ and we get


which simplifies to


and now we'll write the whole thing as one fraction:


and from this, we can see it is useful to write it as


Now simply note that $z^2+z+1>0$ on $|z|<\tfrac12$, so that $\frac{z^2+z+1}{1-4z^2}$ is positive; together with $-4z^2\leq 0$, this means

$$-4z^2\frac{z^2+z+1}{1-4z^2}\leq 0$$

where equality is being reached at $z=0$. Luckily, we also see the maximum of the other term $\frac{9/4}{1-4z^2}$ has a maximum at $z=0$ too (since $1-4z^2\leq 1$ so that $-\tfrac{9/4}{1-4z^2}\leq -\tfrac94$). This means the maximum of $P$ is being found at $z=0$ or $x=\tfrac12$, and $P=-\tfrac94$.

  • $\begingroup$ Unless I am mistaken, $x^2-\frac{3}{4x}-\frac{x}{1-x}$ approaches $-\infty$ for $x \to 1$, and the function has a maximum somewhere in the interior of the interval. – There seems to be an error at $P=x^2+\frac{4x^2+3x-3}{4x(1-x)}$. $\endgroup$ – Martin R Apr 9 '18 at 9:48
  • 1
    $\begingroup$ Yup! Lost a minus sign on the way. Thanks for pointing that out! $\endgroup$ – vrugtehagel Apr 9 '18 at 10:46

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