Question

Let a,b,c be positive real numbers. Prove the inequality


`\displaystyle\\\bf((a+b)^2)/(c) +(c^2)/(a) \geq 4b`

Answer 1

To prove the inequality (a+b)²/c + c²/a ≥ 4b using a, b, c as positive real numbers, we can use the AM-GM inequality.

Step-by-step explanation:

To prove the inequality (a+b)²/c + c²/a ≥ 4b using a, b, c as positive real numbers, we can use the AM-GM inequality. This inequality states that the arithmetic mean (AM) of a set of numbers is always greater than or equal to the geometric mean (GM) of the same set of numbers.

Using the AM-GM inequality, we have:
(a+b)²/c + c²/a = (a²+2ab+b²)/c + c²/a ≥ 2√((a²+2ab+b²)(c²/a)/c) = 2√((a+b)²) = 2(a+b) ≥ 4b

Therefore, the inequality is proven.

Answer 2

Answer: proved

Step-by-step explanation:

Prove the inequality:

`\displaystyle\\((a+b)^2)/(c) +(c^2)/(a) \geq 4b\ \ \ \ \ a > 0\ \ \ \ \ b > 0\ \ \ \ \ c > 0\\`

Multiply both parts of the inequality by ac (a>0, c>0):

`a(a+b)^2+c*c^2\geq 4abc`

Simplify the left side of the inequality using the Cauchy inequality:

`\boxed {t+v\geq 2√(tv) }`

`a(a+b)^2+c(c^2)\geq 2√(a(a+b)^2c(c^2)) \\\\a(a+b)^2+c(c^2)\geq 2(a+b)c√(ac) \\\\a(a+b)^2+c(c^2)\geq 2(ac+bc)√(ac)\ \ \ \ \ (1)`

Simplify the right side of the inequality using the Cauchy inequality:

`ac+bc\geq 2√(acbc) \\\\ac+bc\geq 2√(abc^2) \\\\ac+bc\geq 2c√(ab)`\ \ \ \ \ (2)

Substitute expression (2) into expression (1):

`a(a+b)^2+c(c^2)\geq 2*2c√(ab) √(ab) \\\\a(a+b)^2+c(c^2)\geq4abc`

Hence,

`\displaystyle\\((a+b)^2)/(c) +(c^2)/(a) \geq 4b`