# smallest $k$ s.t. $(x+y)^2\leq k(x^2-xy+y^2)$

I would appreciate if somebody could help me with the following problem

Q: Find $K$?

$$(x+y)^2\leq k(x^2-xy+y^2)$$.

where $\forall x,y\in \mathbb{R}$

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What do you mean 'Find $K$'? I guess you want the least $K$ such that the inequality holds. – xpaul Jul 16 '13 at 13:20
I suppose you are asking for the least $k$ such that the statement is true – Amr Jul 16 '13 at 13:20

You can write the inequality as $$(k-1)x^2-(2+k)xy+(k-1)y^2\ge0$$ and this should hold for every pair $(x,y)$, which means that the discriminant of the polynomial $$(k-1)t^2-(2+k)t+(k-1)$$ must be nonpositive and $k-1>0$. For the discriminant we have $$(2+k)^2-4(k-1)^2\le0$$ which becomes $$-3k^2+12k\le0$$ or $$k^2-4k\ge0$$ that is, $k\le 0$ or $k\ge4$. Taking into account $k>1$, we conclude $k\ge4$. So $4$ is the least value for which the inequality holds on $\mathbb{R}^2$.

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This is part of a general method to study positivity of quadratic forms (sums of terms of degree $2$ in the variables, any number of variables). First I will solve you problem using it and then I will talk a little about the problem in general.

This particular problem:

This $$0\leq(k-1)x^2+(-k-2)xy+(k-1)y^2$$

we can write it as $$(k-1)\left[\left(x^2+\frac{(-k-2)}{(k-1)}xy+\frac{(-k-2)^2y^2}{4(k-1)^2}\right)+(1-\frac{(-k-2)^2}{4(k-1)^2})y^2\right],$$

i.e.

$$(k-1)\left[\left(x+\frac{-k-2}{2(k-1)}y\right)^2+(1-\frac{(-k-2)^2}{4(k-1)^2})y^2\right].$$

We need the positivity of the coefficient of $y^2$.

Therefore $$1\geq\frac{(-k-2)^2}{4(k-1)^2}.$$ From where $$4(k-1)^2\geq(k+2)^2,$$ i.e. $$k(3k-12)\geq0.$$

This gives us $k\leq0$ or $k\geq4$.

The positivity of the coefficient $(k-1)$ gives us $k\geq1$. Intersecting we get $k\geq4$.

In general:

To study the positivity of a quadratic form (this was a quadratic form). You can always bring it to the form $\sum \alpha_iZ_i^2$ for some coefficients $\alpha_i$ and new variables $Z_i$. The form will be positive iff all the coefficients are $\geq0$. There is an algorithm for reducing a quadratic form to this diagonal form. To not explain it in all detail, let me say that it just consists in completing the squares successively for each variable, and when there is only the term $xy$ for some variable $x$ and no $x^2$ you make the substitution $x=Z_1+Z_2$ and $y=Z_1-Z_2$.

From this diagonalization process we get a way to test positivity directly from the coefficients of the original form without having to reduce to the canonical form, see Sylvester's criterion.

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Insert $x=y=1$ to obtain $k\geq 4$. If $k\geq 4$, then $\frac{k+2}{k-1}\leq 2$ and $k-1\geq1$, so: $$k(x^2-xy+y^2)-(x+y)^2 = (k-1)\left(x^2-\frac{k+2}{k-1}xy+y^2\right) \geq x^2-2xy+y^2 = (x-y)^2\geq 0$$

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$$(x+y)^2\leq k(x^2-xy+y^2)\iff (k-1)x^2-(2+k)xy+(k-1)y^2\geq0$$

using the inequality $$xy\leq \frac{1}{2}(x^2+y^2)$$ we have

$$(k-1)x^2-(2+k)xy+(k-1)y^2\geq (k-1)x^2-\frac{2+k}{2}(x^2+y^2)+(k-1)y^2\\ =(\frac{k}{2}-2)(x^2+y^2)\geq 0\iff k\geq 4$$

and if we take $x=y=1$ we find that $4\leq k$.

So $k= 4$ answers the question.

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But is $4$ the smallest one which works? – N. S. Jul 16 '13 at 13:41
Yes $k=4$ is optimal and to see this take $x=y=1$. – user63181 Jul 16 '13 at 13:47
I know, but that is missing from your answer ;) – N. S. Jul 16 '13 at 13:50
Sorry I don't see the word smallest in the title:) – user63181 Jul 16 '13 at 13:54