To prove that 5a + 2b ≥ 6√(ab) for all a, b ≥ 0, we can start by squaring both sides of the inequality:
(5a + 2b)^2 ≥ (6√(ab))^2
Expanding the left side:
25a^2 + 20ab + 4b^2 ≥ 36ab
Rearranging the terms:
25a^2 - 16ab + 4b^2 ≥ 0
Now, let's consider this quadratic expression as a function of a:
f(a) = 25a^2 - 16ab + 4b^2
We need to prove that this quadratic function is always non-negative for all a, b ≥ 0.
Let's find the discriminant of this quadratic function:
Δ = b^2 - 4ac = (-16b)^2 - 4(25)(4b^2) = 256b^2 - 400b^2 = -144b^2
Since b^2 is always non-negative as b ≥ 0, the discriminant Δ is always negative. This implies that the quadratic function f(a) = 25a^2 - 16ab + 4b^2 is always non-negative for all values of a, b ≥ 0.
Therefore, we have proved that 5a + 2b ≥ 6√(ab) for all a, b ≥ 0.
To find when equality holds, we can set the discriminant to be equal to zero:
-144b^2 = 0
This implies that b = 0. Substituting b = 0 into the original inequality, we have:
5a + 2(0) ≥ 6√(a(0))
5a ≥ 0
So, equality holds when a ≥ 0 and b = 0.
In conclusion, the inequality 5a + 2b ≥ 6√(ab) holds for all a, b ≥ 0, and equality holds when a ≥ 0 and b = 0.
Prove that 5a + 2b ≥ 6√(ab) for all a,b ≥ 0 and find when equality holds
3 answers
do this again without squaring both sides
To prove 5a + 2b ≥ 6√(ab) for all a, b ≥ 0 without squaring both sides, let's start by assuming a and b are non-negative real numbers.
By the Arithmetic Mean-Geometric Mean inequality, we have:
(5a + 2b)/2 ≥ √(5a * 2b)
Simplifying this inequality, we get:
5a + 2b ≥ 2√(10ab)
Now, we need to show that 2√(10ab) ≥ 6√(ab). We can divide both sides of the inequality by √(ab) since a and b are non-negative:
2√(10ab)/√(ab) ≥ 6√(ab)/√(ab)
Simplifying further:
2√(10) ≥ 6
√(10) ≥ 3
Since √(10) is approximately 3.1623, we see that the inequality √(10) ≥ 3 is true. Therefore, we have:
2√(10ab) ≥ 6√(ab)
Combining this inequality with the previous one, we obtain:
5a + 2b ≥ 2√(10ab) ≥ 6√(ab)
Hence, we have proven that 5a + 2b ≥ 6√(ab) for all a, b ≥ 0.
To find when equality holds, we note that equality in the Arithmetic Mean-Geometric Mean inequality occurs when all the terms are equal. In this case, equality holds when 5a = 2b and 5a = 2√(5a * 2b). By solving these two equations simultaneously, we find that equality holds when a = 0 and b = 0.
Therefore, the inequality 5a + 2b ≥ 6√(ab) holds for all a, b ≥ 0, and equality holds only when a = 0 and b = 0.
By the Arithmetic Mean-Geometric Mean inequality, we have:
(5a + 2b)/2 ≥ √(5a * 2b)
Simplifying this inequality, we get:
5a + 2b ≥ 2√(10ab)
Now, we need to show that 2√(10ab) ≥ 6√(ab). We can divide both sides of the inequality by √(ab) since a and b are non-negative:
2√(10ab)/√(ab) ≥ 6√(ab)/√(ab)
Simplifying further:
2√(10) ≥ 6
√(10) ≥ 3
Since √(10) is approximately 3.1623, we see that the inequality √(10) ≥ 3 is true. Therefore, we have:
2√(10ab) ≥ 6√(ab)
Combining this inequality with the previous one, we obtain:
5a + 2b ≥ 2√(10ab) ≥ 6√(ab)
Hence, we have proven that 5a + 2b ≥ 6√(ab) for all a, b ≥ 0.
To find when equality holds, we note that equality in the Arithmetic Mean-Geometric Mean inequality occurs when all the terms are equal. In this case, equality holds when 5a = 2b and 5a = 2√(5a * 2b). By solving these two equations simultaneously, we find that equality holds when a = 0 and b = 0.
Therefore, the inequality 5a + 2b ≥ 6√(ab) holds for all a, b ≥ 0, and equality holds only when a = 0 and b = 0.