If a^x=b^y=c^z,where x,y,z are unequal positive number and a,b,c are in G.P.,then x³+z³ is? (Answer is >2y³ but how?)

Hi Swati,
Please find below the solution to the asked query:

$$\begin{gathered} {\mathrm{a}}^{\mathrm{x}}={\mathrm{b}}^{\mathrm{y}}={\mathrm{c}}^{\mathrm{z}}=\mathrm{k} \\ \Rightarrow \mathrm{a}={\mathrm{k}}^{\frac{1}{\mathrm{x}}}, \mathrm{b}={\mathrm{k}}^{\frac{1}{\mathrm{y}}}, \mathrm{c}={\mathrm{k}}^{\frac{1}{\mathrm{z}}} \\ \mathrm{As} \mathrm{a},\mathrm{b},\mathrm{c} \mathrm{are} \mathrm{in} \mathrm{G}.\mathrm{P}. \\ \Rightarrow \frac{\mathrm{b}}{\mathrm{c}}=\frac{\mathrm{c}}{\mathrm{b}} \\ \Rightarrow \frac{{\mathrm{k}}^{\frac{1}{\mathrm{y}}}}{{\mathrm{k}}^{\frac{1}{\mathrm{x}}}}=\frac{{\mathrm{k}}^{\frac{1}{\mathrm{z}}}}{{\mathrm{k}}^{\frac{1}{\mathrm{y}}}} \\ \Rightarrow \frac{1}{\mathrm{y}}-\frac{1}{\mathrm{x}}=\frac{1}{\mathrm{z}}-\frac{1}{\mathrm{y}} \\ \Rightarrow \frac{1}{\mathrm{x}}\text{'}\frac{1}{\mathrm{y}},\frac{1}{\mathrm{z}} \mathrm{are} \mathrm{in} \mathrm{A}.\mathrm{P}. \\ \Rightarrow \mathrm{x},\mathrm{y},\mathrm{z} \mathrm{are} \mathrm{in} \mathrm{H}.\mathrm{P}. \\ \mathrm{y}=\mathrm{H}.\mathrm{M}. \mathrm{of} \mathrm{x},\mathrm{y} \\ \mathrm{We} \mathrm{know} \mathrm{that} \mathrm{for} \mathrm{positive} \mathrm{numbers}: \\ \mathrm{A}.\mathrm{M}.>\mathrm{G}.\mathrm{M}.>\mathrm{H}.\mathrm{M} \\ \Rightarrow \sqrt{\mathrm{xz}}>\mathrm{y} ;\mathrm{condition}\left(\mathrm{i}\right) \\ \mathrm{Now} \mathrm{consider} \mathrm{two} \mathrm{numbers} {\mathrm{x}}^3 \mathrm{and} {\mathrm{z}}^3. \\ \mathrm{A}.\mathrm{M}.>\mathrm{G}.\mathrm{M} \\ \Rightarrow \frac{{\mathrm{x}}^3+{\mathrm{z}}^3}{2}>\sqrt{{\mathrm{x}}^3{\mathrm{z}}^3} \\ \Rightarrow \frac{{\mathrm{x}}^3+{\mathrm{z}}^3}{2}>{\left(\sqrt{\mathrm{xz}}\right)}^3 \\ \mathrm{Using} \mathrm{condition}\left(\mathrm{i}\right), \mathrm{we} \mathrm{get}, \\ \Rightarrow \frac{{\mathrm{x}}^3+{\mathrm{z}}^3}{2}>{\mathrm{y}}^3 \\ \mathbf{\Rightarrow }{\mathbf{x}}^{\mathbf{3}}\mathbf{+}{\mathbf{z}}^{\mathbf{3}}\mathbf{>}\mathbf{2}{\mathbf{y}}^{\mathbf{3}}\mathbf{ }\left(\mathbf{Answer}\right) \end{gathered}$$

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