Let x and y be positive reals with x + y = 1. Which are correct? A: xy has minimum 1/4. B: 1/x + 1/y has minimum 4. C: x² + y² has minimum 1/2. D: √x + √y has maximum 2.

Step-by-step solution

1. 1

   Set up the one degree of freedom. With $x+y=1$ and both positive, write $x=t$, $y=1-t$ with $0<t<1$. Every quantity below is a function of $t$ alone, and by symmetry the extreme values will typically occur either at $t=\tfrac12$ or in the limits $t\to 0^{+}$, $t\to 1^{-}$.

2. 2

   A is wrong — 1/4 is the maximum of xy. By AM–GM, $xy\le\left(\dfrac{x+y}{2}\right)^{2}=\dfrac14$, with equality at $x=y=\tfrac12$. As $t\to 0^{+}$, $xy=t(1-t)\to 0$, so $xy$ has no positive minimum; its infimum is $0$. Incorrect.

3. 3

   B is correct. Since $x+y=1$,

   $\frac1x+\frac1y=\frac{x+y}{xy}=\frac{1}{xy}$

   and $xy\le\tfrac14$ from the previous step, so $\dfrac{1}{xy}\ge 4$, with equality at $x=y=\tfrac12$. Correct.

4. 4

   C is correct. Using $x^{2}+y^{2}=(x+y)^{2}-2xy=1-2xy$ and $xy\le\tfrac14$,

   $x^{2}+y^{2}\ge 1-2\left(\frac14\right)=\frac12$

   with equality again at $x=y=\tfrac12$. Correct.

5. 5

   D is wrong — the maximum is √2, not 2. Squaring,

   $\left(\sqrt x+\sqrt y\right)^{2}=x+y+2\sqrt{xy}=1+2\sqrt{xy}\le 1+2\left(\frac12\right)=2$

   so $\sqrt x+\sqrt y\le\sqrt2\approx 1.414$, attained at $x=y=\tfrac12$. The value $2$ is never reached. Incorrect.

6. 6

   Conclude and spot-check.

   $\boxed{\text{B and C}}$

   At $x=y=\tfrac12$: $xy=0.25$, $\tfrac1x+\tfrac1y=4$, $x^{2}+y^{2}=0.5$, $\sqrt x+\sqrt y=1.414$. At $x=0.1$, $y=0.9$: $xy=0.09$, $\tfrac1x+\tfrac1y=11.11>4$, $x^{2}+y^{2}=0.82>0.5$, $\sqrt x+\sqrt y=1.265<1.414$ — every claim above behaves as stated.