## Show the inequality \$ sec x ge 1+ (x ^ 2/2) \$

Test inequality

$$sec (x) ge 1+ (x ^ 2/2)$$

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## partial derivative – Ge the canonical form of the equation

Dice $$x ^ 2u_ {xx} + 2xyu_ {xy} – 3y ^ 2u_ {yy} – 2xu_x + 4yu_y + 16x ^ 4u = 0$$
I have to find the canonical form. The determinant is the following $$D = 16x ^ 2y ^ 2$$
I considered cases when $$x$$ Y $$and$$ can be $$0$$. When calculating for the remaining case (where both $$x$$ Y $$and$$ It is not $$0$$ and the type is hyperbolic), I'm not getting its canonical form. I'm receiving this
$$(- 3y ^ 2 – 6x ^ 3y + 9x ^ 6) u _ { xi xi} + (-3y ^ 2 + frac {2y} {x} + frac {1} {x ^ 2} ) u _ {eta} + (-6y ^ 2 – 6x ^ 2 + frac {2y (1-3x ^ 4)} x) u _ { eta xi} + ( frac {-4} {x} + 4y) u_ eta + 4yu_ xi + 16x ^ 4u = 0$$
But the point is that this does not meet the requirements for the canonical form of hyperbolic type.
Any help would be appreciated.

## precalculus algebra: show the inequality \$ sum x + 6 ge 2 ( sum sqrt {xy}) \$

Leave $$x; Y; z in R ^ +$$ such that $$x + y + z + 2 = xyz$$. Such that $$x + y + z + 6 ge 2 ( sqrt {xy} + sqrt {yz} + sqrt {xz})$$

This inequality is not homogeneous and look at the condition that I thought
I would replace the variables $$x; Y; z$$ such that

+)$$x ^ 2 + y ^ 2 + z ^ 2 + 2xyz = 1$$. Leave $$x = frac {2a} { sqrt { left (a + b right) left (a + c right)}}$$

+)$$xy + yz + xz + xyz = 4$$. Leave $$a = frac {2 sqrt {xy}} { sqrt { left (y + z right) left (x + z right)}}$$.Leave $$x = frac {2a} {b + c}$$

but failed. Please explain to me how I can obtain this substitution (if I have a solution by substitution)

I also tried to solve it by $$u, v, w$$.Leave $$sum_ {cyc} x = 3u; sum_ {cyc} xy; Pi_ {cyc} a = w ^ 3 (3u + 2 = w ^ 3; u, v, w> 0)$$ so $$u le w ^ 3-3u$$ or $$4u le w ^ 3$$ but stagnant (I'm very bad at $$uvw$$)

## functional analysis – \$ conv (A) = { sum_ {i = 1} ^ n lambda _i x_i: sum_ {i = 1} ^ n lambda_i = 1, lambda_i ge 0, x_i in A, i = 1, cdots, n } \$

Leave $$X$$ be a normed vector space and $$A$$ be a subset of $$X$$. $$conv (A)$$
It is called the intersection of all convex subsets of $$X$$ that contain
$$A$$

a) Show that conv (A) is a convex set

b) show that

$$conv (A) = { sum_ {i = 1} ^ n lambda _i x_i: sum_ {i = 1} ^ n lambda_i = 1, lambda_i ge 0, x_i in A, i = 1, cdots, n }$$

c) yes $$A$$ it is compact then it is $$conv (A)$$ compact?

d) Show that if $$A subseteq mathbb {R} ^ n$$ it is compact then $$conv (A)$$ is
compact

a) Take two elements $$a$$ Y $$b$$ at the intersection of all convex subsets of $$X$$ that contain $$A$$. Now take $$lambda a + (1- lambda) b$$. It is contained in each of the subsets of $$X$$ that contain $$A$$, therefore it is contained in the intersection. Q.E.D.

second)

$$leftarrow$$:

Suppose by induction that $$sum_ {i = 1} ^ n lambda_i x_i$$ It belongs to conv (A). We must prove that $$sum_ {i = 1} ^ {k + 1} lambda_i x_i$$, with $$sum_ {i = 1} ^ {k + 1} = 1$$ It also belongs.

$$sum_ {i = 1} ^ {k + 1} lambda_i x_i = sum_ {i = 1} ^ {k} lambda_i x_i + lambda_ {k + 1} x_ {k + 1} = sum_ {i = 1} ^ {k} lambda_i x_i + (1- sum_ {i = 1} ^ n lambda_i) x_ {k + 1}$$

Now choose $$delta$$ such that $$delta sum_ {i = 1} ^ {k} lambda_i = 1$$, so

$$frac { delta} { delta} left ( sum_ {i = 1} ^ {k} lambda_i x_i + (1- sum_ {i = 1} ^ n lambda_i) x_ {k + 1 } right = = left ( frac {1} { delta} sum_ {i = 1} ^ {k} ( delta lambda_i) x_i + frac {1} { delta} ( delta-1) x_ {k + 1} right) = \ frac {1} { delta} x + left (1- frac {1} { delta} right) x_ {k + 1}$$

which is a collection of elements of $$A$$ that adds up to $$1$$

$$rightarrow$$:

I can only say that $$x in conv (A)$$, so $$x = 1x + 0 cdot all$$ Therefore, it is a combination that adds to $$1$$ of elements of $$A$$?

by do), give me a clue: show that conv (A union B) is the image through a continuous function of the compact $${( alpha, beta; alpha, beta ge 0, alpha + beta = 1) } times A times B$$. I do not understand how to show this.

re) Some clue?

## Probability: anti-concentration: upper limit for \$ P ( sup_ {a in mathbb S_ {n-1}} sum_ {i = 1} ^ na_i ^ 2Z_i ^ 2 ge epsilon) \$

Leave $$mathbb S_ {n-1}$$ be the unitary sphere in $$mathbb R ^ n$$ Y $$z_1, ldots, z_n$$ be a sample i.i.d of $$mathcal N (0, 1)$$.

Dice $$epsilon> 0$$ (It can be assumed that it is very small), which is reasonable upper limit for the tail probability $$P ( sup_ {a in mathbb S_ {n-1}} sum_ {i = 1} ^ na_i ^ 2z_i ^ 2 ge epsilon)$$ ?

• Using ideas from this other answer (MO link), you can set the not uniform Limit of anti-concentration: $$P ( sum_ {i = 1} ^ na_i ^ 2z_i ^ 2 le epsilon) le sqrt {e epsilon}$$ for all $$a in mathbb S_ {n-1}$$.

• The uniform analog is another story. Can coverage numbers be used?

## \$ Π (x) ge log log x \$ holds \$ 2 le x e e {{e ^ 3} <5.3 times 10 ^ 8 \$?

The book Theory of Numbers by G H Hardy, et al. test $$π (x) ge log log x$$ for $$x> e ^ {e ^ 3}$$. There is a way to try out that also goes for $$2 le x e e ^ {e ^ 3}$$ otherwise (in the worst case) some valuable source to verify this numerically?