Prove that: $a^2+b^2+(1-a-b)^2\ge \frac {1}{3}$ Where $a$ and $b$ are any given real number. I have tried solving it using partial derivative.
$$
s=a^2+b^2+(1-a-b)^2$$
$$\frac{\partial s}{\partial a}=2a-2(1-a-b) \tag{1}$$
$$\frac{\partial s}{\partial b}=2b-2(1-a-b) \tag{2}$$
for maxima both (1) and (2) are 0..from here we get two equations from where we get values of $a$ and $b$
$$2a-2(1-a-b)=0 \tag{3}$$
$$2b-2(1-a-b)=0 \tag{4}$$
putting the values of $a$ and $b$ we find from (3) and (4) in the function the maximum value of the function should be $\frac {1}{3}$.which in this case is not

You want to estimate $R=a^2+b^2+c^2$, where $a+b+c=1$. Note that $R$ is the squared length of the radius vector to $(a,b,c)$. The equation $a+b+c=1$ gives you some plane, and the shortest radius vector to the plane is the perpendicular on it. It's easy to calculate that its length(height of a pyramid) is equal to $\frac{1}{ \sqrt{3}}$, so $R\ge\frac 1 3$.
Anyway, I like the way with Cauchy-Schwarz inequality more - it is the most obvious and easy way.
