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
 A: Hint: Apply Cauchy-Schwarz inequality to the vectors
$$
\vec{u}=(1,1,1)\qquad\text{and}\qquad\vec{v}=(a,b,1-a-b).
$$

Hint: Look at your equations $(3)$ and $(4)$. Can you deduce that $a=b$ at a critical point? Can you then solve for $a$ and $b$?
A: Using Titu Andreescu's Lemma, we have
$\dfrac{a^2}{1}+\dfrac{b^2}{1}+\dfrac{(1-a-b)^2}{1} \ge \dfrac{(a+b+1-a-b)^2}{3}=\dfrac{1}{3}$
A: 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
$$\forall\,a,b\in\Bbb R\;,\;\;a^2+b^2+(1-a-b)^2\ge\frac29+\left(1-\frac23\right)^2=\frac29+\frac19=\frac13\;\ldots$$
A: One more approach. We start with a bit simpler inequality:
$$a^2+(1-a)^2 > \frac{1}{2},$$
which can be proven without words by the following picture:
$\hspace{100pt}$
(if you insist on explanation: you want to make dark gray as small as possible, but clearly as long as $a > b$ then the red part is bigger than green).
Now, if there are three squares (as in the question), and some two are of different sizes, then you can make them of equal size without changing the third square, but at the same time decreasing the dark gray area. This process converges to three squares of the same size and gives an intuition why the result holds ;-)
$\hspace{100pt}$
I know it's not a proper proof, but I hope it helps ;-)
A: 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 $\dfrac{1}{3}$
A: 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.
