# Tag Info

### Distance of ellipse to the origin

You can always try to write it in polar coordinates: $$3r^2+4r^2\sin\theta\cos\theta=20$$ $$r^2=\frac{20}{3+2\sin2\theta}\ge\frac{20}{3+2}=4$$
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Accepted

### Distance of ellipse to the origin

Hint: $4xy \le 2(x^2+y^2), 3x^2+3y^2 = 3(x^2+y^2) \implies 5(x^2+y^2) \ge 20 \implies x^2+y^2 \ge .....$
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Accepted

• 52.2k

### Minimizing $x^2+y^2+z^2$ subject to $xy -z + 1 = 0$ via Lagrange multipliers

$$x^2+y^2+z^2=x^2+y^2+z^2+2(xy-z+1)=(x+y)^2+(z-1)^2+1\geq1.$$ The equality occurs for $x=y=0$ and $z=1$, which says that we got a minimal value.

### Distance of ellipse to the origin

By inspection (or see this for additional reference if required), we can see that this is an ellipse with centre at origin and axes of symmetry $y=\pm x$. Substituting into the ellipse equation gives ...
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When you solve the system with Lagrange method, your variables are $x,y,z$ and $\lambda$. One solution to the system is $$(x,y,z,\lambda)=(1,0,0,\frac{1}{4}),$$ which matches your intuitive solution. ...
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### If $x$ and $y$ are real number such that $x^2+2xy-y^2=6$, then find the minimum value of $(x^2+y^2)^2$

Use $$2(x^2+y^2)^2-(x^2+2xy-y^2)^2=(x^2-2xy-y^2)^2\geq0.$$

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### Showing that $x^{\top}Ax$ is maximized at $\max \lambda(A)$ for symmetric $A$
There is a cleaner proof altogether that circumvents the need to consider the square root of $A$. Consider the spectral decomposition of $A$ given by $A = V\Lambda V^{\top}$. For some \$x \in \mathbb{R}...