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inequality
Daytuz   1
N Mar 23, 2025 by alexheinis
Consider the function \( f \) defined on \( \mathbb{R}^2 \) by
\[f(x, y) = x^4 + y^4 - 2(x - y)^2.\]
Show that there exist \( (\alpha, \beta) \in \mathbb{R}^2 \) (and determine them) such that
\[\forall (x, y) \in \mathbb{R}^2, f(x, y) \geq \alpha \| (x, y) \|^2 + \beta,\]where \( \| \cdot \| \) denotes the Euclidean norm.
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Daytuz
Mar 23, 2025
alexheinis
Mar 23, 2025
inequality
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Daytuz
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Consider the function \( f \) defined on \( \mathbb{R}^2 \) by
\[f(x, y) = x^4 + y^4 - 2(x - y)^2.\]
Show that there exist \( (\alpha, \beta) \in \mathbb{R}^2 \) (and determine them) such that
\[\forall (x, y) \in \mathbb{R}^2, f(x, y) \geq \alpha \| (x, y) \|^2 + \beta,\]where \( \| \cdot \| \) denotes the Euclidean norm.
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alexheinis
10508 posts
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Let's consider $f$ on the circle $x^2+y^2=r$ and write $p:=xy$. Then $f(x,y)=(x^2+y^2)^2-4p^2-2(r-2p)=r^2-2r-4(p^2+p)$. Also $|p|\le r/2$ hence $\min(f)=r^2-2r-4(r^2/4+r/2)=-4r$. It follows that we want $-4r\ge ar+b$ for all $r\ge 0$ hence $(a+4)r\le -b$.
Taking $r=0$ we have $b\le 0$ and with $r\rightarrow \infty$ we find $a\le -4$. Hence all pairs with $a\le -4, b\le 0$.
The best inequality is $f(x,y)\ge -4(x^2+y^2)$.
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