<s>[INST] equation $\displaystyle a_{n+1}.x^2-2x\sqrt{a^2_{1}+a^2_{2}+.......+a^2_{n}+a^2_{n+1}}+\left(a_{1}+a_{2}+.......+a_{n}\right) = 0$ has real roots The Largest positive integer $n$ such that for all real no. $a_{1},a_{2},.......,a_{n},a_{n+1}.$ The equation 
$\displaystyle a_{n+1}.x^2-2x\sqrt{a^2_{1}+a^2_{2}+.......+a^2_{n}+a^2_{n+1}}+\left(a_{1}+a_{2}+.......+a_{n}\right) = 0$ has real roots.
My Try:: If Given equation has real roots. Then its Discriminant $\geq 0$
So $\displaystyle 4\left(a^2_{1}+a^2_{2}+.......+a^2_{n}+a^2_{n+1}\right)-4.a_{n+1}.\left(a_{1}+a_{2}+.......+a_{n}\right)\geq 0$
So $\left(a^2_{1}+a^2_{2}+.......+a^2_{n}+a^2_{n+1}\right)-a_{n+1}.\left(a_{1}+a_{2}+.......+a_{n}\right)\geq 0$
after that How can i solve it,plz explain me  Thanks
 [/INST] Let $x=a_{n+1}$ then the last inequality is $x^2-xT+S \ge 0$ where $T=a_1+a_2+\cdots+a_n$ and $S=a_1^2+a_2^2+\cdots+a_n^2$. This requires its own  discriminant $\Delta=T^2-4s\le 0$. As you try increasing the number of $a_i$'s you see at 4 elements the inequality can be made to work but not after that. Here is how it turns out after a bit of algebra
for $n=1$ we have
$-\Delta=3a_1^2\ge 0$.
for $n=2$ we have 
$-\Delta=2(a_1^2+a_2^2)+(a_1-a_2)^2 \ge 0$.
for $n=3$ we have 
$-\Delta=1(a_1^2+a_2^2+a_3^2)+(a_1-a_2)^2 +(a_2-a_3)^2+(a_3-a_1)^2\ge 0$.
for $n=4$ we have 
$-\Delta=0(a_1^2+a_2^2+a_3^2+a_4^2) \\ +(a_1-a_2)^2+(a_1-a_3)^2+(a_1-a_4)^2 +(a_2-a_3)^2+(a_2-a_4)^2+(a_3-a_4)^2\ge 0$.
for $n=5$ the inequality does not work for all possible $a_i$'s.  Use the suggestion of @vonbrand, above, of setting all of them equal to say $t$ to see $-\Delta=4(5t^2)-(5t)^2=-5t^2 <0 $ for $t\ne 0$.
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