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

LHS$ =a^2+b^2+1-2(a+b)+(a+b)^2 \ge \dfrac{(a+b)^2}{2}+1-2(a+b)+(a+b)^2=\dfrac{3}{2}(a+b)^2-2(a+b)+1=\left(\sqrt{\dfrac{3}{2}}(a+b)-\sqrt{\dfrac{2}{3}}\right)^2+\dfrac{1}{3} \ge \dfrac{1}{3}  $
first "=" is $a=b$,2nd "=" is $a+b=\dfrac{2}{3} \to a=b=\dfrac{1}{3}$ to get $\dfrac{1}{3}$
