Find the minimum value of $f$ on $\{(a,b,c)\in \mathbb{R^3}\mid a+b+c=1\}$ where $f(a,b,c)=\int_{0}^1 (a+bx+cx^2)^2dx$. Find the minimum value of $f$ on $\{(a,b,c)\in \mathbb{R^3}\mid a+b+c=1\}$ where $f(a,b,c)=\int_0^1 (a+bx+cx^2)^2 \, dx$.
What I did is 
$$
\int_0^1 (a+bx+cx^2)^2 \, dx=a^2+a \left( b+\frac{2c}{3} \right) + \frac{b^2}{3} + \frac{bc}{2}+\frac{c^2}{5}.
$$
Then we consider the function $F(a,b,c)=a^2+a(b+\frac{2c}{3})+\frac{b^2}{3} + \frac{bc}{2} + \frac{c^2}{5}-\lambda(a+b+c-1)$. Thus, we can consider $\nabla F=0$ with $a+b+c=1$ and try to find $a,b$ and $c$.
However, since the function and the constraint seem to be related, I am curious whether there is some other way to see the minimum without computing Lagrange multiplier. Thank you.

Considering a quadric
$$x^2+\frac{y^2}{3}+\frac{z^2}{5}+\frac{yz}{2}+\frac{2zx}{3}+xy=\lambda$$
which touches the plane $x+y+z=1$ at $(a,b,c)$.
The tangent plane will be
$$ax+\frac{by}{3}+\frac{cz}{5}+
\frac{cy+bz}{4}+\frac{az+cx}{3}+\frac{bx+ay}{2}=\lambda$$
Comparing coefficients:
$$a+\frac{b}{2}+\frac{c}{3}=
\frac{a}{2}+\frac{b}{3}+\frac{c}{4}=
\frac{a}{3}+\frac{b}{4}+\frac{c}{5}=\lambda$$
gives $$(a,b,c)=3\lambda(1,-8,10)$$
and $$\lambda=\frac{1}{9}$$
