Ask a New Question
Search
given real numbers x, a, b with x>= a>= b>= 0, show that sqrt x+b - sqrt x-a >= sqrt x+a - sqrt x-b
Ask a New Question
or
answer this question
.
Similar Questions
given real numbers x, a, b with x>= a>= b>= 0, show that sqrt x+b - sqrt x-a >= sqrt x+a - sqrt x-b
2 answers
Let $x$ and $y$ be nonnegative real numbers. Find the smallest real number $k$ such that
\[\sqrt{x} + \sqrt{y} \le k \sqrt{x +
1 answer
Prove: [1/sqrt(2)] [sqrt(a) + sqrt(b)] <= sqrt(a + b) <= sqrt(a) + sqrt(b) for all non-negative real numbers a and b.
1 answer
f(x)= 4-x^2 and g(x)= sqrt (x)
find the implied domain of fg(x) fg(x)= f(sqrt(x)) fg(x)= 4-(sqrt(x))^2 fg(x)=4-(sqrt x)(sqrt x)
2 answers
more similar questions