a, b, c are the sides of $△ A B C$ such that a+b+c=2.,
find the max value of $a^{2} + b^{2} + c^{2} + 2 a b c < 2$

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Solution

Given, , $a + b + c = 2.$
$1 - a + 1 - b + 1 - c = 1$
$x + y + z = 1$ ;
where $x = 1 - a, y = 1 - b, z = 1 - c$ .
Since, $a + b > c$ $\Rightarrow 0 < c < 1$
Similarly, $0, < a, b < 1,$ hence $0 < x, y, z < 1.$
Now, $a^{2} + b^{2} + c^{2} + 2 a b c = {(1 - x)}^{2} + {(1 - y)}^{2} + 2 (1 - x) (1 - y) (1 - z)$
$= 3 - 2 (x + y + z) + (x^{2} + y^{2} + z^{2}) + 2 [1 - (x + y + z) + (x y + y z + z x) - x y z]$
$= 1 + x^{2} + y^{2} + z^{2} - 2 x y z + 2 (x y + y z + z x)$
$= 1 + {(x + y + z)}^{2} - 2 x y z$
$= 2 - 2 x y z < 2$ ; as $0 < x, y, z < 1$
$a^{2} + b^{2} + c^{2} + 2 a b c < 2$

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