Any ideas?

Up! Up!

I think finding $z$ from first equation and putting this in the second one + case bash solves problem but bump for synthetic sol. Synthetic, I mean not finding 1 variable as values of other one, just using inequalities without a lot of bash. Since it's P1 in TsT I think there will be such solution.

bump......

socrates wrote: Let $x, y, z$ be real numbers such that $$\displaystyle{\begin{cases} x^2+y^2+z^2+(x+y+z)^2=9 \\ xyz \leq \frac{15}{32} \end{cases}} $$Find the maximum possible value of $x.$ I got $\max{x}=\frac{5}{2}$, which occurs for $(x,y,z)=\left(\frac{5}{2},\frac{\sqrt{13}-5}{8},\frac{-5-\sqrt{13}}{8}\right).$

Deleted....

For $(x,y,z) = (3,0,0)$, the given two conditions are not even satisfied.

Bump/////////////

socrates wrote: Let $x, y, z$ be real numbers such that $$\displaystyle{\begin{cases} x^2+y^2+z^2+(x+y+z)^2=9 \\ xyz \leq \frac{15}{32} \end{cases}} $$Find the maximum possible value of $x.$ $1.Case: x \le 0 \rightarrow x_{max} \le 0.$ $2.Case: x > 0 $ $$x^2+y^2+z^2+(x+y+z)^2=9 \iff x\left(x+2y+2z\right)^2+4\left(\frac{15}{32}-xyz \right)+\left(x-\frac{5}{2} \right)\left(3x^2+\frac{15x}{2}+\frac{9}{4}\right)=0,(*)$$ $$(*) \rightarrow x \le \frac{5}{2}, \left(x=\frac{5}{2} \iff y+z=-\frac{5}{4} ; yz=\frac{3}{16}\rightarrow y=\frac{-5+\sqrt{13}}{8} ; z=\frac{-5-\sqrt{13}}{8}\right)$$$$\rightarrow \boxed {x_{max}=\frac{5}{2}}.$$ .

Another version of the solution (maybe a bit more motivated): We express everything in terms of $y+z$ and $yz$. So if $x>0$ (as we may assume), the second condition becomes $yz \le \frac{15}{32x}$ and the first condition becomes \[2(y+z)^2+2x(y+z)-2yz=9-2x^2.\]Now for each value of $x$, we get an equation between $y+z$ and $yz$ and an inequality on $yz$ and we can ask whether we can find such $y,z$. There are two possible obstructions, namely that we cannot even find $yz$ and $y+z$ or that we can find $yz$ and $y+z$, but there is no real solution for $y,z$ afterwards. Experimenting, it turns out that the first will be the limiting case here (which is convenient for us!). So we should just use the inequality on $yz$ to get \[2(y+z)^2+2x(y+z) \le 9-2x^2+\frac{15}{16x}.\]To see whether this will have a solution for $y+z$, we can complete the square and get \[(2(y+z)+x)^2 \le 18-3x^2+\frac{15}{8x}.\]So what we have proved is that $18-3x^2+\frac{15}{8x} \ge 0$. From here it is easy to factorize this to \[(2x-5)(12x^2+30x+3) \le 0\]and hence obtain that indeed $x \le \frac{5}{2}$. Btw, I think that there is a typo in the solution in #10 where it should be $\frac{3}{4}$ instead of $\frac{9}{4}$ in the last term of the identity.

Hm....,tough problem

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