5.0 Basic properties

  Inequalities

• If $a>b>0$ and $n>0$ then ${a^n} > {b^n}$ and ${a^{\frac{1}{n}}} > {b^{\frac{1}{n}}}$.
• If $a>b>0$ and $x>0$ then ${a^x} > {b^x}$.
• If $x>y>0$ and $a>1$ then ${a^x} > {a^y}$.
• If $x>y>0$ and $0<a<1$ then ${a^x} < {a^y}$.
• If $x>y>0$ and $a>1$ then ${\log _a}x > {\log _a}y$.
• If $x>y>0$ and $0<a<1$ then ${\log _a}x < {\log _a}y$.

Question 5. If $a$, $b$, $c$ are positive and not all equal then prove that $$(a + b + c)(bc + ca + ab) > 9abc$$

Solution: We have $$(a + b + c)(bc + ca + ab) - 9abc$$

$$\begin{equation} \begin{aligned} = abc + {a^2}c + {a^2}b + {b^2}c + abc + {b^2}a + {c^2}b + {c^2}a + abc - 9abc \\ = a({b^2} + {c^2}) + b({c^2} + {a^2}) + c({a^2} + {b^2}) - 6abc \\ = a({b^2} + {c^2} - 2bc) + b({c^2} + {a^2} - 2ca) + c({a^2} + {b^2} - 2ab) \\ = a{(b - c)^2} + b{(c - a)^2} + c{(a - b)^2} \\\end{aligned} \end{equation} $$

which is positive because each term is positive. Thus, $$\begin{equation} \begin{aligned} (a + b + c)(bc + ca + ab) - 9abc > 0 \\ \Rightarrow (a + b + c)(bc + ca + ab) > 9abc \\\end{aligned} \end{equation} $$

Question 6. If $a$, $b$, $c$ be positive and unequal , prove that $$\frac{{b + c}}{a} + \frac{{c + a}}{b} + \frac{{a + b}}{c} > 6$$ or $$bc(b + c) + ca(c + a) + ab(a + b) > 6abc$$

Solution: Since Arithmetic mean > Geometric mean $$\begin{equation} \begin{aligned} \frac{{\frac{a}{b} + \frac{b}{a}}}{2} > \sqrt {\frac{a}{b}.\frac{b}{a}} \\ \frac{a}{b} + \frac{b}{a} > 2 \\\end{aligned} \end{equation} $$

Similarly, $$\frac{b}{c} + \frac{c}{b} > 2$$ and $$\frac{c}{a} + \frac{a}{c} > 2$$

Adding the three equations, we get $$\begin{equation} \begin{aligned} \left( {\frac{a}{b} + \frac{b}{a}} \right) + \left( {\frac{b}{c} + \frac{c}{b}} \right) + \left( {\frac{c}{a} + \frac{a}{c}} \right) > 6 \\ \left( {\frac{b}{a} + \frac{c}{a}} \right) + \left( {\frac{c}{b} + \frac{a}{b}} \right) + \left( {\frac{a}{c} + \frac{b}{c}} \right) > 6 \\ \frac{{b + c}}{a} + \frac{{c + a}}{b} + \frac{{a + b}}{c} > 6{\text{ or}} \\ bc(b + c) + ca(c + a) + ab(a + b) > 6abc \\\end{aligned} \end{equation} $$

Question 7. Solve for $x$:

(a) ${\log _3}(x - 4) > 1$

(b) ${\log _3}({\log _{\frac{1}{2}}}({\log _4}(x)) > 0$

Solution: (a) $$\begin{equation} \begin{aligned} {\log _3}(x - 4) > 1 \\ {\log _3}(x - 4) > {\log _3}3 \\ x - 4 > 3 \\ x > 7 \\ \Rightarrow x \in \left( {7,\infty } \right) \\\end{aligned} \end{equation} $$

(b) $$\begin{equation} \begin{aligned} {\log _3}({\log _{\frac{1}{2}}}({\log _4}(x)) > 0 \\ {\log _3}({\log _{\frac{1}{2}}}({\log _4}(x)) > {\log _3}1 \\ {\log _{\frac{1}{2}}}({\log _4}(x)) > 1 \\ {\log _4}(x) > \frac{1}{2} \\ {\log _4}(x) > {\log _4}{4^{\frac{1}{2}}} \\ x > {4^{\frac{1}{2}}} \\ x > 2 \\ \Rightarrow x \in \left( {2,\infty } \right) \\\end{aligned} \end{equation} $$