parmenides51 wrote: Let a$,b,c,d$ and $x,y,z,t$ be real numbers such that $0\le a,b,c,d \le 1$ , $x,y,z,t \ge 1$ and $a+b+c+d +x+y+z+t=8$. Prove that $a^2+b^2+c^2+d^2+x^2+y^2+z^2+t^2\le 28$ Let $$L(a,b,c,d,x,y,z,t) = a^2+b^2+c^2+d^2+x^2+y^2+z^2+t^2 $$First of all, we see that: $$L(a,b,c,d,x,y,z,t) \leq L(0,0,0,0,x+a,y+b,z+c,t+d)$$Therefore, it is enough to check the case $a=b=c=d=0$. In this case, we have $x,y,z,t \geq 1$ with $x+y+z+t=8$ and we need to maximise: $$L(x,y,z,t) = x^2+y^2+z^2+t^2$$WLOG, let's assume $x \geq y \geq z \geq t$. We have: $$L(x-t+1,y,z,1)-L(x,y,z,t) = 2(x-1)(t-1) \geq 0 $$Therefore, we conclude that: $$L(x,y,z,t) \leq L(x+t-1,y,z,1) \leq L(5,1,1,1) = 28$$Therfore: $$\boxed{L_{max} = 28}$$achieved at $a=b=c=d=0$, $y=z=t=1$, $x=5$.

parmenides51 wrote: Let a$,b,c,d$ and $x,y,z,t$ be real numbers such that $0\le a,b,c,d \le 1$ , $x,y,z,t \ge 1$ and $a+b+c+d +x+y+z+t=8$. Prove that $a^2+b^2+c^2+d^2+x^2+y^2+z^2+t^2\le 28$ Using Am-Gm inequality, we have: $$x^2+y^2+z^2+t^2+a^2+b^2+c^2+d^2 \geq \frac{1}{8}(x+y+z+t+a+b+c+d)^2=8$$Therefore: $$\boxed{L_{min} = 8}$$achieved at $a=b=c=d=x=y=z=t=1$.

First, notice that $0 \le a,b,c,d \le 1$ and $1 \le x,y,z,t \le 5$. So, we have \begin{align*}a^2+b^2+c^2+d^2+x^2+y^2+z^2+t^2 &\le a+b+c+d+(6x-5)+(6y-5)+(6z-5)+(6t-5) \\&\le 6(a+b+c+d+x+y+z+t)-20 \\&= 28.\end{align*}

NaPrai wrote: First, notice that $0 \le a,b,c,d \le 1$ and $1 \le x,y,z,t \le 5$. So, we have \begin{align*}a^2+b^2+c^2+d^2+x^2+y^2+z^2+t^2 &\le a+b+c+d+(6x-5)+(6y-5)+(6z-5)+(6t-5) \\&\le 6(a+b+c+d+x+y+z+t)-20 \\&= 28.\end{align*} Very nice.

Thank you all for your splendid participation and to mister Takis for posting. I added something to it, just for the collection.

mihaig wrote: Thank you all for your splendid participation and to mister Takis for posting. I added something to it, just for the collection. Wow amazing. I would be grateful if you could post a solution.

minageus wrote: Wow amazing. I would be grateful if you could post a solution. Thank you. I could post a solution, but my strong belief is that you and the other friends from here should try it first. Again, it's my duty to inform all the participants that the new problem is very difficult.

Ok this is my attempt: $a+b+c+d+x+y+z+t=8\Leftrightarrow (a+b+c+d+x+y+z+t)^2=64\Leftrightarrow \sum_{cyc}a^2+\sum_{cyc}x^2=64-2(a+b+c+d)(x+y+z+t)$. So the given inequality takes this new form: $64-2(a+b+c+d)(x+y+z+t)+20\sqrt{abcdxyzt}\leq 28$. Therefore we now have that: $(a+b+c+d)(x+y+z+t)\geq 2+10\sqrt{abcdxyzt}$. Here I have some ideas. I will edit the post when I will be able to reach more close to the solution.

Beautiful. I like your dedication.

Can someone post the solution to the latter problem please?

Please, leave it for a while. Eventually, I'll post my solution. But it's not the right moment for me yet.

Ok, I just thought that you forgot it and I thought it would be nice if I made the thread alive.

Since $0\leq a,b,c,d\leq 1$, we have \begin{align*}a^2+b^2+c^2+d^2+x^2+y^2+z^2+t^2 &\leq a+b+c+d+x^2+y^2+t^2 \\ &\leq 28 \iff x^2+y^2+z^2+t^2\leq 20+x+y+z+t.\end{align*}Now, WLOG, assume that $x\leq y\leq z\leq t$, make the change $x=x'+1,y=y'+1,z=z'+1,t=t'+1$ where $x',y',z',t'\geq 0$, so it's suffices to prove that \[x'^2+y'^2+z'^2+t'^2+x'+y'+z'+t'\leq 20.\]Observe that $0\leq 8-(x+y+z+t) \implies x'+y'+z'+t'\leq 4$ and so \begin{align*}&x'^2+y'^2+z'^2+t'^2+x'+y'+z'+t' \leq 20 \\ &\iff x'^2+y'^2+z'^2+t'^2=(x'+y'+z'+t')^2-\sum_{sym}x'y'\leq 16 \\ &\iff x'=y'=z'=0, t'=4\end{align*}hence equality holds at $a=b=c=d=0, x=y=z=1, t=5. \blacksquare$

At first we claim 1.$x+y+z+t=S \leq 8$ 2.$xy+yz+zt+tx+yt+zx+6 \geq 3S$ 3.$a^2 + b^2 + c^2 + d^2 \leq a+b+c+d$ So we have: $a^2 +b^2 +c^2 +d^2 +x^2 +y^2 +z^2 +t^2 \leq a+b+c+d+x^2 +y^2 +z^2 +t^2 = S^2 -S -2(xy+yz+zt+tx+yt+zx)+8 \leq S^2 -7S +20 = (S-8)(S+1)+28 \leq 28$

$a^2+b^2+c^2+d^2+x^2+y^2+z^2+t^2 \le a+b+c+d+ 2(x+y+z+t) - 4+(x-1)^2 + (y-1)^2 + (z-1)^2 + (t-1)^2 \le x+y+z+t+4+(x+y+z+t - 4)^2 \le 4+8+(8-4)^2 = 28$ With equality if $(a,b,c,d,x,y,z,t) = (0,0,0,0,1,1,1,5)$

We can write $a=1-a_1 ...$ $x=1+x_1...$ and result is trivial