<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] Using Titu Andreescu's Lemma, we have
$\dfrac{a^2}{1}+\dfrac{b^2}{1}+\dfrac{(1-a-b)^2}{1} \ge \dfrac{(a+b+1-a-b)^2}{3}=\dfrac{1}{3}$
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