MathsGee is Zero-Rated (You do not need data to access) on: Telkom |Dimension Data | Rain | MWEB

0 like 0 dislike
15 views
Let $a, b, c$ be positive real numbers. Show that
$$\frac{a^{2}}{b^{2}}+\frac{b^{2}}{c^{2}}+\frac{c^{2}}{a^{2}} \geq \frac{b}{a}+\frac{c}{b}+\frac{a}{c}$$
| 15 views

0 like 0 dislike
Applying AM-GM on $\frac{a^{2}}{b^{2}}$ and $\frac{b^{2}}{c^{2}}$, we obtain
$$\frac{a^{2}}{b^{2}}+\frac{b^{2}}{c^{2}} \geq 2 \sqrt{\frac{a^{2} b^{2}}{b^{2} c^{2}}}=2 \frac{a}{c} .$$
Observe that this last term is one of the terms on the RHS of our inequality, which is a great news for us. Similarly, we have
$$\begin{array}{l} \frac{b^{2}}{c^{2}}+\frac{c^{2}}{a^{2}} \geq 2 \frac{b}{a} \\ \frac{c^{2}}{a^{2}}+\frac{a^{2}}{b^{2}} \geq 2 \frac{c}{b} \end{array}$$
Adding up these 3 inequalities and dividing by 2 , we get that
$$\frac{a^{2}}{b^{2}}+\frac{b^{2}}{c^{2}}+\frac{c^{2}}{a^{2}} \geq \frac{b}{a}+\frac{c}{b}+\frac{a}{c} \cdot \square$$
by Diamond (75,914 points)

0 like 0 dislike