Suppose $\chi$ is a nonzero, non-trivial character of $G$ and $\chi(g)$ is a non-negative real number for all $g \in G$. Prove that $\chi$ is reducible.
My try : I want to prove $\langle\chi,\chi\rangle>1$ then we are done.
Now, $$\langle\chi,\chi\rangle=\frac 1{|G|}\sum_{g \in G}\chi(g)\overline{\chi(g)}=\frac 1{|G|}\sum_{g \in G}\chi(g)^2>0$$ How do I know that $\sum_{g \in G}\chi(g)^2>|G|$?
Any help?
One note : The chapter that I am reading didn't prove Schur's Orthogonality yet so I don't want to overkill the problem. I don't know whether the solution will actually involve the steps of Schur's Orthogonality relation or not!
If so please mention that then I will look into this later. I just came to know the name after seeing the other related solutions.
I presume you are working over $\mathbb C$.
$\chi(1)$ must be positive, so if $\chi(g)$ is real and nonnegative for all $g$, then $d=\frac{1}{|G|}\sum \chi(g)$ is a positive real. Now if a representation $V$ affords $\chi$, then $d$ is exactly the $\mathbb C$-dimension of $V^G$. But $V^G$ is a $G$-submodule of $V$ on which $G$ acts trivially. Hence $V$ has a positive-dimensional submodule on which $G$ acts trivially. The conclusion is that $V$ is reducible or trivial.
If you permit yourself to use orthogonality, the argument is easier: $\langle \chi, \chi_1\rangle$ is positive, so $\chi$ is reducible or $\chi=\chi_1$ is trivial.