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How do I prove that $$(xy+2)^2+(x-1)^2+(y-1)^2 > 1$$ for all real $x,y$? I tried setting the partial derivative to $0$ but I couldn't get it to work.

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    (1) Get the partials but don't expand. (2) Set them equal to zero and subtract the equations. (3) You should see how to factor the new equation. That should get you a little further. – 2017-02-14
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    Thanks, Ill see how it goes. – 2017-02-14
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    If I sent you off into partial derivative land, sorry -- that was a false lead. – 2017-02-14

5 Answers 5

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I came to my senses -- no need for Calculus.

I see that my solution is similar to Michael Burr's, but anyway, here it is:

Let $f(x,y) = (xy + 2)^2 + (x-1)^2 + (y-1)^2$.

If $x < 0$ then ($x-1)^2 > 1$, so $f(x,y) > 1$.

If $y < 0$ then ($y-1)^2 > 1$, so $f(x,y) > 1$.

If $x,y \ge 0$ then $(xy + 2)^2 \ge 4$, so $f(x,y) > 1$.

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    So very slick. +1 – 2017-02-14
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Let $\,p=xy\,$ and $\,s=x+y\,$, then the LHS is:

$$ \begin{align} (xy+2)^2 + (x-1)^2+(y-1)^2 & = (p+2)^2 + (x^2+y^2) - 2(x+y) + 2 \\ & = (p^2 + 4p + 4)+(s^2-2p) -2s + 2 \\ & = p^2+2p+s^2-2s+6 \\ & = (p+1)^2 +(s-1)^2+ 4 \\ & \ge 4 \end{align} $$

Equality is attained for $p=-1,s=1$ i.e. when $x,y$ are the roots of $t^2-t-1=0\,$.

Of course, the LHS being $\ge 4$ implies that it's $\gt 1\,$.

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    That gets the actual minimum. Nice. – 2017-02-14
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    @quasi Thanks. Actually, the first posted answer gave me the idea that "*if that argument works, then there must be a tighter bound for sure*". – 2017-02-14
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    Well done! Your Change-of-variables approach is most convincing ... not to mention the goodie "Locates & quantifies the minimum." One of the upvotes is mine ;-) – 2017-02-14
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If the sum of the three squares is less than or equal to $1$, then each of the squares must, individually, be less than or equal to $1$. For $(x-1)^2\leq1$, it must be that $0\leq x\leq2$. Similarly, for $(y-1)^2\leq1$, then $0\leq y\leq2$. Now, look at $xy+2$, the smallest this could be is $2$, but this leads to a LHS greater than $1$.

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Because $$(xy+2)^2+(x-1)^2+(y-1)^2 - 1 = {\frac { \left( x{y}^{2}+x+2\,y-1 \right) ^{2}+3(1+y^2)+ \left( {y}^{2}-y-1 \right) ^{2}}{{y}^{2}+1}}.$$

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    Upvote—and, or but, I encourage you to add some text wrapping your insightful equality (I did assert it using SymPy ). Pretty sure it'll be helpful & more inviting to be upvoted. – 2017-02-14
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Using @quasi 's original comment, consider the function $$ f(x,y)=(xy+2)^2+(x-1)^2+(y-1)^2. $$ If we take partial derivatives of $f$, we get \begin{align*} \frac{\partial f}{\partial x}&=2(xy+2)y+2(x-1)\\ \frac{\partial f}{\partial y}&=2(xy+2)x+2(y-1). \end{align*} At minima, these are equal to zero, so we consider \begin{align*} 2(xy+2)y+2(x-1)&=0\\ 2(xy+2)x+2(y-1)&=0. \end{align*} Next, we can subtract the second equation from the first equation to get $$ 2(xy+2)(y-x)+2(x-y)=0. $$ Factoring this, we have $$ 2(y-x)(xy+2-1)=0 $$ or that $$ 2(y-x)(xy+1)=0. $$ Therefore, minima occur when, either 1. $x=y$ or 2. $xy+1=0$.

In the first case, $(xy+2)^2$ simplifies to $(x^2+2)^2$, since squares are always positive, the inside of the square is at least $2$, so the LHS is at least $4$.

In the second case, if $xy+1=0$, then $xy=-1$, and, plugging this into the formula for $f$, we see that the LHS is at least $(xy+2)^2=(-1+2)^2=1$. Now, we can't have both $x=1$ and $y=1$ and have their product $-1$, so at least one of $(x-1)^2$ or $(y-1)^2$ is positive, and the LHS is greater than $1$.