set $x=2\sin{a},y=2\cos{a},z=3\sin{b},t=3\cos{b},a,b\in [0,\pi]$ $6\leq xt+yz=6\sin{a}\cos{b}+6\cos{b}\sin{a}\leq 6\sin{(a+b)}\leq 6$ so $a+b=\frac{\pi}{2}$ $z+x=3\sin{b}+2\sin{a}=3\sin{(\frac{\pi}{2}-a)}+2\sin{a}=3\cos{a}+2\sin{a}\leq\sqrt{3^2+2^2}=\sqrt{13}$

and $\min{(z+x)}=-\sqrt{13}$

better latex tenplusten wrote: $\boxed{A2}$ Find the maximum value of $z+x$ if $(x,y,z,t)$ satisfy the conditions $ \begin{cases} x^2+y^2=4 \\ z^2+t^2=9 \\ xt+yz\geq 6 \end{cases}$

Doesn't $(x,y,z,t)=(0,2,3,0)$ work and give a larger value of $x^2+z^2=9$?

Firstly notice that $(xt+yz)^2-36\geq 0$ And we have $(x^2+y^2)(z^2+t^2)=36$ From here we get $-(xz-yt)^2\geq 0$ or $xz=yt$. This is useful because we see that: $x^2+y^2+z^2+t^2=(x+z)^2+(y-t)^2=13$ Now we have that $x+z\leq \sqrt{13}$

Well, you can get with Cauchy that xt+yz=6.

@Steve12345 nice solution but, can you give me such x, y, z, t satisfying the 2 first conditions, x+z=sqrt13 and xt+yz=6???

Because I tried, but I cant find any, while for max(x+z)= 5/sqrt2 (which is only 0.07 smaller than sqrt13) I have found numbers satisfying it...

x = 4/(sqrt13) y = 6/(sqrt13) t = 6/(sqrt13) z = 9/(sqrt13) Should satisfy the equations and have the max value of (x + z) .

That is the solution i got.

Attachments:

And can someone explain how do i write Latex here ?

--- From the given conditions, we have: \[ (x^2 + y^2)(z^2 + t^2) = 36 = (xt + yz)^2 + (xz - yt)^2 \geq 36 + (xz - yt)^2, \]which implies that \( xz - yt = 0 \). Now, it is clear that: \[ x^2 + z^2 + y^2 + t^2 = (x + z)^2 + (y - t)^2 = 13. \] The maximum value of \( z + x \) is \( \sqrt{13} \), and it is achieved when: \[ x = \sqrt{\frac{4}{13}}, \quad y = t = \sqrt{\frac{6}{13}}, \quad z = \sqrt{\frac{9}{13}}. \] ---