Maximum and minimum value of $2xy+2yz+7xz$

I hate to do this, but I feel inclined to bash using Lagrange multipliers. For any $x^2+y^2+z^2=k$, $0\le k\le100$, we want $\left<2y+7z,2x+2z,2y+7x\right>=\lambda\left<2x,2y,2z\right>$. Solving for $\lambda$, we get $2y+7z=2x\lambda$, $2x+2z=2y\lambda$, and $2y+7x=2z\lambda$. Eliminate $x$ by substituting $x=y\lambda-z$ to get $2y+7z=2y\lambda^2-2z\lambda$ and $2y+7(y\lambda-z)=2z\lambda$. Plugging in the second as $2z\lambda$, we get the quadratic $0=2y\lambda^2-7y\lambda-4y=y(2\lambda+1)(\lambda-4)$, so $\lambda=-\frac{1}{2}\text{ or }4$, or $y=0$. Whew, $\lambda$ is somewhat easy to work with (thanks Jorge!).

In the case that $y=0$, we have $7xz$ to optimize. Assume WLOG that $x,z\ge0$, so $7xz\le7\left(\sqrt{\frac{x^2+z^2}{2}}\right)^2\le7\frac{100}{2}=350$. That means $y=0$ gives a maximum of $350$ ($x=z=5\sqrt{2}$) and minimum of $-350$ ($x=5\sqrt{2},z=-5\sqrt{2}$).

In the case that $\lambda=4$, $2y+7z=4(2x)$ and $2x+2z=4(2y)$. Solving for $y$ and $z$, $y=\frac{1}{2}x$ and $z=x$, which plugs in as $2xy+2yz+5zx=9x^2$ and $x^2+y^2+z^2=\frac{9}{4}x^2\le100$. The minimum here is $0$ ($x=y=z=0$) clearly, but the maximum is $9\frac{4}{9}100=400$ ($x=z=\frac{20}{3},y=\frac{10}{3}$).

In the case that $\lambda=-\frac{1}{2}$, $2y+7z=-\frac{1}{2}(2x)$ and $2x+2z=-\frac{1}{2}(2y)$. Solving for $y$ and $z$, we get $y=-4x$ and $z=x$, which plugs in as $2xy+2yz+5zx=-11x^2$ and $x^2+y^2+z^2=18x^2\le100$. The maximum here is clearly $0$ ($x=y=z=0$), but the minimum is $-4\frac{100}{18}=-\frac{550}{9}$ ($x=z=\frac{5\sqrt{2}}{3},y=-\frac{20\sqrt{2}}{3}$).

In effect, the maximum value is $\boxed{400}$ at $\left(\frac{20}{3},\frac{10}{3},\frac{20}{3}\right)$, and the minimum value is $\boxed{-350}$ at $\left(5\sqrt{2},0,-5\sqrt{2}\right)$.

Now the real question is what is the non-LM solution?

If we want the greatest possible value of $2xy+2yz+2xz$ (instead of $2xy+2yz+7xz$ ) we can use the inequalities $2xy\leq x^2+y^2$, $2yz\leq y^2+z^2$ and $2xz\leq x^2+z^2$, in this case the maximum occurs when $x=y=z$.

But in our problem we can try to use similar inequalities like $2xy\leq Ax^2+By^2$, where $A$ and $B$ are suitable constants.

For minimum: $2xy+2yz+7xz = (x+y+z)^2 - (x^2+y^2+z^2) + \frac{5}{2} (x+z)^2 - \frac{5}{2}(x ^2 + z^2) \\ \ge - (x^2+y^2+z^2) - \frac{5}{2}(x ^2 + z^2) \ge -100 - \frac{5}{2} (100) = -350.$ The equality holds when $x+y+z=0, x+z=0, x^2+y^2+z^2=100, x^2+z^2=100$. There is an obvious solution, that is $y=0, x=-z, x^2=z^2=50$.

Let $k = 2xy+2yz+ 7zx$. Then $k \leq \dfrac{100( 2xy+2yz+ 7zx)}{x^2+y^2+z^2}$, which becomes $\begin{aligned} kx^2-x(200y+700z)+ky^2-200yz+kz^2\leq 0. \end{aligned}$ Since the solutions of $kx^2-x(200y+700z)+ky^2-200yz+kz^2 = 0$ are real (solutions are $x$ and $\dfrac{200y+700z}{k}-x$), its discriminant $\Delta_x$ is nonnegative: $\begin{aligned} \Delta_x = (200y+700z)^2-4k(ky^2-200yz+kz^2) &\geq 0\\ \implies (40000-4k^2)y^2 + (280000+800k)yz+(490000-4k^2)z^2&\geq 0.\qquad(\star) \end{aligned}$ If $z = 0$, then $(\star)$ becomes $(40000-4k^2)y^2 \geq 0$ and since $y^2\geq 0$ for all $y\in\mathbb{R}$ we must have $40000-4k^2\geq 0\implies |k|\leq 100$. If $z\neq 0$, then dividing both sides of $(\star)$ and taking $t=\dfrac{y}{z}$ gives $\begin{aligned} (40000-4k^2)t^2 + (280000+800k)t+(490000-4k^2)\geq 0. \end{aligned}$ Since the roots of the equation $(40000-4k^2)t^2 + (280000+800k)t+(490000-4k^2)=0$ are real (note that $t = \dfrac{y}{z}\in\mathbb{R}$, its discriminant $\Delta_t$ is nonnegative: $\begin{aligned} \Delta_t = (280000+800k)^2-4(40000-4k^2)(490000-4k^2)\geq 0\\ \implies k^3-142500k-7000000\leq 0. \end{aligned}$ (I used Wolfram Alpha to obtain this cubic inequality). Since the roots of $k^3-142500k-7000000=0$ are $k=-350, -50, 400$ (also obtained from Wolfram Alpha), the minimum value of $k = 2xy+2yz+7zx$ is $\boxed{-350}$ at $(x,y,z)=(-5\sqrt{2}, 0, 5\sqrt{2})$ and the maximum value is $\boxed{400}$ at $(x,y,z)=\left(\dfrac{20}{3}, \dfrac{10}{3}, \dfrac{20}{3}\right)$.

For maximum: $2xy+2yz+7xz = 4(x^2+y^2+z^2) - \frac{1}{2}((x-2y)^2+(z-2y)^2 + 7(x-z)^2) \le 400.$ The equality holds when $x=2y=z$ and $x^2+y^2+z^2=100$.

Or apply Cauchy-Schwarz to:

$(2xy+2yz+7xz)^2 = [(x)(2y) + (2y)(z) + (x)(z) + (x)(z)+ (x)(z) + (z)(x) + (z)(x)+(z)(x)+(z)(x)]^2\\ \le[x^2+(2y)^2+x^2+x^2+x^2+z^2+z^2+z^2+z^2][(2y)^2+z^2+z^2+z^2+z^2+x^2+x^2+x^2+x^2]\\ =(4x^2+4y^2+4z^2)^2$

Or simply as Jorge pointed out: $2xy = x(2y)\le \frac{1}{2}(x^2+4y^2), 2yz = (2y)z\le \frac{1}{2}(z^2+4y^2), 7xz \le \frac{7}{2}(x^2+z^2)$ then add all three together.

Why has the diagram of a sphere been given?

I have no idea what to do here.I tried establishing a bound but I am not sure what to use here.I have tried to use some elementary inequalities such as Cauchy-Schwarz and AM-GM.Using AM-GM,we find that

$\dfrac{x^2+y^2+z^2}{3}\ge [(xy)(yz)(zx)]^\dfrac{1}{3}\le \dfrac{xy+yz+zx}{3}$

But I am not sure how that helps us.Also,

$x^2+y^2+z^2\ge \dfrac{(x+y+z)^2}{3}$

by Titu's lemma.I have also tried to rewrite the expression as

$2xy+2yz+7zx=2(xy+yz+zx)+5zx$

and was wondering whether I can apply AM-GM here.

how can i become as intelligent as you