Calculus logarithm function

Question

I was reading a Calculus Book. In the chapter of Logarithm function I found this exercise but I was not able to complete it myself so I came to you:

Prove that for any $x,y>0$ it is true that $$ {1\over x} + {1\over y} \ge {4\over {x+y}}$$

Use the inequality to prove that if $a,b,c>1$ then it is true that $$ \log_a(bc) + \log_b(ca) + \log_c(ab) \ge 4[\log_{bc}(a) + \log_{ca}(b) + \log_{ab}(c)]$$

Answer 1

For your first question: it is equivalent to $$\frac{x+y}{xy}\geq \frac{4}{x+y}$$ or $$(x+y)^2\geq 4xy$$ or $$(x-y)^2\geq 0$$ and your second one is equivalent to $$\log_{a}{bc}+\log_{b}{ca}+\log_{c}{ab}\geq 4\left(\frac{1}{\log_{a}{bc}}+\frac{1}{\log_{b}{ca}}+\frac{1}{\log_{c}{ab}}\right)$$

Answer 2

If $x,y>0$ this means that $x+y>0$ and $xy>0$ as well. Look at the given inequality

$$\frac{1}{x}+\frac{1}{y} \ge \frac{4}{x+y}$$

$$\frac{x+y}{xy} \ge \frac{4}{x+y}$$

$$(x+y)^2 \ge 4xy$$

Continue to prove this equation, which is identical to the given one for the conditions you have stated.