Mistake in another approach for the old problem

Question

In the question being discussed here Prove that $|f(x)| \le \frac{3}{2}$ when $f(x)=ax^2+bx+c$ i get it how they used the triangle inequality and the bound here that is this : $$|f(x)|=|(\frac{f(1)}{2})(x^2+x)+(\frac{f(-1)}{2})(x^2-x)+(f(0))(1-x^2)|$$

$$\le |\frac{x^2+x}{2}|+|\frac{x^2-x}{2}|+|1-x^2|$$ $\leq5/4$ but i would like to know why this method too dont give the same maximum value as the above gives ? My method : $$|f(x)|$$ = |($$\frac{f(1)+f(-1)-2f(0)}{2} )x^2 + (\frac{f(1)-f(-1)}{2}) x + (f(0))| \leq \frac{1+1-2(-1)}{2} x^2 + \frac{1-(-1)}{2} |x| + 1$$ $$\leq 2x^2 + |x| + 1$$ whose maximum in that domain is even greater than $3/2$ , so whats wrong with this approach ? Triangle inequality like this we cannot apply ? Why so if thats the case?

Answer 1

You correctly identified the greatest possible value of $a,$ the greatest possible value of $b,$ and the greatest possible value of $c.$

What you have not accounted for is the fact that these greatest possible values cannot all occur simultaneously. The greatest $a$ requires $f(−1)=1$ but the greatest $b$ requires $f(−1) = -1.$ The greatest $a$ requires $f(0)=-1$ but the greatest $c$ requires $f(0)=1.$

These facts do not invalidate what you wrote; what you wrote is still true. But some ways of setting an upper bound are "tighter" and others are "looser". The answer to the question was a method with a tighter bound than yours.

Answer 2

What you have done is correct. It's just that $|f(x)|\le 2x^2+|x|+1$ does not mean that $|f(x)|\le 3/2$ is ruled out. If you had proved, for instance $|f(x)|\le 2$ that too would be correct.