Answer:
Now considering that
for all x : [tex]x^{2} \geq 0[/tex]
we show that
[tex]ab + bc + ac \leq a^{2} + b^{2} + c^{2}\\2ab + 2bc + 2ac \leq 2a^{2} + 2b^{2} + 2c^{2}\\\\0 \leq 2a^{2} + 2b^{2} + 2c^{2} -2ab -2bc -2ac \\\\\\0 \leq (a^{2} + b^{2} - 2ab) + (a^{2} + c^{2} - 2ac) + (b^{2} + c^{2} - 2bc) \\\\0 \leq (a+b)^2 + (a+c)^2 + (b+c)^2 \\\\[/tex]
Step-by-step explanation:
