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How to prove the set $S=\{(x,y,z)\in \mathbb R^3|z\geq 0,x^2+y^2\leq z^2\}$ is convex?

So to prove this I took $(x,y,z),(x_1,y_1,z_1)\in S$ and $\lambda\in (0,1).$

Then $x^2+y^2\leq z^2$ and $x_1^2+y_1^2\leq z_1^2$ and $z,z_1\geq 0.$

Now to prove $\{\lambda x+(1-\lambda)x_1\}^2+\{\lambda y+(1-\lambda)y_1\}^2\leq \{\lambda z+(1-\lambda)z_1\}^2$ , it's coming like $\lambda^2(x^2+y^2-z^2)+(1-\lambda)^2(x_1^2+y_1^2-z_1^2)+2\lambda(1-\lambda)(xx_1+yy_1-zz_1)\leq 0.$

So from here, I was trying to prove $(xx_1+yy_1-zz_1)\leq 0$ because the former terms are already negative in the above inequality.

I'm not sure whether $(xx_1+yy_1-zz_1)\leq 0$ is true or not?

Can we prove the inequality $xx_1+yy_1-zz_1\leq 0$? Or there is another way to prove the convexity of the set $S$?

Any help is appreciated. Thank you.

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Yes, you can show the inequality $$x x_1+y y_1 \leq z z_1.$$ It is sufficient to show the inequality $$(x x_1 + y y_1)^2 \leq (z z_1)^2.$$ (Note that sufficiency follows from the fact that $z, z_1 \geq 0$.) To show the squared form, multiply the inequalities $$x^2 + y^2 \leq z^2$$ $$x_1^2 + y_1^2 \leq z_1^2$$ together. Then use the inequality $a^2 + b^2 \geq 2ab$ with for $a=x y_1$ and $ b= x_1y$ . This gives $$z^2 z_1^2 \geq (x^2+y^2)(x_1^2+y_1^2)=(xx_1)^2+(xy_1)^2+(x_1 y)^2+(y y_1)^2 \geq (x x_1)2+2(x x_1y y_1)+(y y_1)2=(x x_1+y y_1)2.$$

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Note that $(x,y,z)\in S\implies \lambda (x,y,z)\in S$ since $$(\lambda x)^2+(\lambda y)^2=\lambda^2(x^2+y^2)\le \lambda^2z^2=(\lambda z)^2$$

Thus $S$ is a cone.

Prove that a cone is convex if and only if it is closed under addition.

Then you see that your condition $xx_1+yy_1\le zz_1$ is the closure under addition of $(x,y,z)$ and $(x_1,y_1,z_1)$.

This one is obtained using $ab\le \frac {a^2+b^2}2$ since both $\begin{cases}x^2+y^2\le z^2\\{x_1}^2+{y_1}^2\le{z_1}^2\end{cases}$ are verified for points of $S$.

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Here we have $$ xx_1 + yy_1 \le \sqrt{(x^2+y^2)(x_1^2+y_1^2)} \le \sqrt{z^2}\sqrt{z_1^2}= zz_1\,. $$ The final equality relies on $z,z_1 \ge 0$.

The first inequality is an example of the Cauchy-Schwarz inequality, and can be directly shown in this case. (Note that the inequality is trivial if $xx_1+yy_1 < 0$.)

$$ x^2y_1^2 - 2xyx_1y_1 + y^2x_1^2 = (xy_1-yx_1)^2 \ge 0 \\ x^2y_1^2 + y^2x_1^2 \ge 2xyx_1y_1 \\ x^2x_1^2+ x^2y_1^2 + y^2x_1^2 + y^2y_1^2 \ge x^2x_1^2+ 2xyx_1y_1 + y^2y_1^2 \\ (x^2+y^2)(x_1^2+y_1^2) \ge (xx_1+yy_1)^2 $$

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Let us consider the usual norm $\|\vec a\|=\sqrt{a_1^2+a_2^2}$ on $\mathbb R^2$

Notice also that the condition describing $S$ is exactly $\sqrt{x^2+y^2} \le z$, so $$S=\{(x,y,z); z\ge0, \|(x,y)\|\le z\}.$$ So now if we have two points $(x_1,y_1,z_1),(x_2,y_2,z_2)\in S$, then we get for any $\lambda\in(0,1)$ $$\|(\lambda x_1+(1-\lambda)x_2,\lambda y_1+(1-\lambda)y_2)\| \overset{(*)}\le \lambda\|(x_1,y_1)\| + (1-\lambda)\|(x_2,y_2)\| \le \lambda z_1+(1-\lambda)z_2.$$

The inequality $(*)$ follows from triangle inequality applied to vectors $(\lambda x_1,\lambda y_1)$ and $((1-\lambda)x_2,(1-\lambda)y_2)$. But we can view it also as convexity of the function $(x,y)\mapsto\|(x,y)\|$. See, for example Why is every $p$-norm convex? and other posts linked there.

We have shown that $$\|(\lambda x_1+(1-\lambda)x_2,\lambda y_1+(1-\lambda)y_2)\| \le \lambda z_1+(1-\lambda)z_2,$$ which means that also the convex combination $(\lambda x_1+(1-\lambda)x_2,\lambda y_1+(1-\lambda)y_2,\lambda z_1+(1-\lambda)z_2)$ belongs to $S$.

Notice that similar approach would work to prove convexity of $S=\{(x_1,\dots,x_{n+1}); x_{n+1}\ge 0, \sum_{k=1}^n x_k^2 \le x_{n+1}^2\}$ in $\mathbb R^{n+1}$. (This condition can be equivalently stated as $\|(x_1,\dots,x_n)\|\le\sqrt{x_{n+1}}$.)

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