Do you mean that

1. a = √(b² + c²) or a = √(b² + c²) = (b² + c²)¹⁄² but not a = √b² + √c² or a = (b²)¹⁄² +( c²)¹⁄²

2. If a² = b² + c², then a ≠ b + c (not possible)

\(\displaystyle (a + b)^2 = (a + b)(a + b) = a(a + b) + b(a + b) \implies\)

\(\displaystyle (a + b)^2 = a^2 + ab + ba + b^2 = a^2 + ab + ab + b^2 = a^2 + 2ab + b^2.\)

Any questions so far?

\(\displaystyle a \ne 0 \text { and } b \ne 0 \implies ab \ne 0 \implies 2ab \ne 0 \implies\)

\(\displaystyle a^2 + 2ab + b^2 \ne a^2 + b^2 \implies (a + b)^2 \ne a^2 + b^2.\)

Still following?

\(\displaystyle \therefore a \ne 0 \text { and } b \ne 0 \implies \sqrt{(a + b)^2} \ne \sqrt{a^2 + b^2}\)

\(\displaystyle \text {THUS, } a + b \ne \sqrt{a^2 + b^2} \text { if } a \ne 0 \text { and } b \ne 0.\)

EDIT: To expand a bit on what Dr. Peterson wrote.

You can prove, for any pair of real numbers, the truth of:

\(\displaystyle a = 0 \text { or } b = 0 \implies \sqrt{a^2 + b^2} = |a| + |b|.\)

It is false however that, for all real numbers a and b,

\(\displaystyle a = 0 \text { or } b = 0 \implies \sqrt{a^2 + b^2} = a + b.\)