<s>[INST] 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
 [/INST] Taking it from where you left it:
$$a=1-a-b\implies a=\frac{1-b}2\\
b=1-a-b\implies b=\frac{1-a}2$$
So substituting the first eq. into the second one and using the symmetry of both:
$$b=\frac{1-\frac{1-b}2}2=\frac{1+b}4\implies b=\frac13\;\;\text{, and thus also}\;\;a=\frac13$$
Continuing with the partial derivatives
$$\frac{\partial^2s}{\partial a^2}=4=\frac{\partial^2s}{\partial b^2}\;\;,\;\;\frac{\partial^2s}{\partial a\partial b}=\frac{\partial^2s}{\partial b\partial a}=2$$
$$\text{Thus, at}\;\;\left(\frac13\,,\,\frac13\right)\;,\;\;\text{the Hessian is positive definite and we get a minimum there.}$$
From here, we get that
$$\forall\,a,b\in\Bbb R\;,\;\;a^2+b^2+(1-a-b)^2\ge\frac29+\left(1-\frac23\right)^2=\frac29+\frac19=\frac13\;\ldots$$
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