For all nonnegative real numbers a, b, and c, if a^2 + b^2 = c^ 2 , then a + b ≥ c Plz Help with the solution as I can not find the answer.?

I imagine you wish to prove this.

Let

d = c - (a + b)

We know that:

a^2+b^2 = c^2

Let's rewrite c as (a+b+d)

We have:

a^2+b^2 = (a+b+d)^2

Expanding on the right side:

a^2+b^2 = a^2 + ab + ad + ab + b^2 + bd + ad + bd + d^2

Subtracting a^2+b^2 from both sides we have:

0 = 2ab + 2ad + 2bd +d^2

-2ab = 2ad + 2bd +d^2

as a and b are non-negative we have

0 >= 2d(a+b) + d^2

Now if a and be both equal 0, we know c=0 and we are done (because a^2+b^2 = 0^2+0^2 = 0 = c). Otherwise, we may divide by (a+b) and get:

0>=2d + d^2/(a+b)

As d^2 must also be non-negative we get:

0>=d

Reassigning d we get:

0>= c - (a + b)

So

(a+b) >= c

qed.

a^2 + b^2 = c^ 2

=>4^2 + 3^2 = 5^ 2

=>16+9=25

=>25=25

So, a=4,b= 3 and c=5