UPSC CDS 1 2022 Math Paper

Concept:The given inequality simplifies to $(x^2 - y^2)^2 > 0$, which is always true when $x \neq y$ (both non-zero).Explanation:Start with $\frac{x^6 + y^6}{x^4 + y^4} > \frac{x^4 + y^4}{x^2 + y^2}$ (given $x \neq y \neq 0$).Cross-multiply (all denominators are positive because squares are non‑negative and $x,y \neq 0$):$(x^6 + y^6)(x^2 + y^2) > (x^4 + y^4)(x^4 + y^4)$.Expand both sides: $x^8 + x^6 y^2 + x^2 y^6 + y^8 > x^8 + 2x^4 y^4 + y^8$.Cancel $x^8$ and $y^8$ from both sides: $x^6 y^2 + x^2 y^6 > 2x^4 y^4$.Factor left side: $x^2 y^2 (x^4 + y^4) > 2x^4 y^4$.Divide both sides by $x^2 y^2$ (non‑zero): $x^4 + y^4 > 2x^2 y^2$.Rearrange: $x^4 + y^4 - 2x^2 y^2 > 0$, i.e. $(x^2 - y^2)^2 > 0$.This inequality holds for every real $x$ and $y$ with $x \neq y$, irrespective of whether $x > y$ or $x < y$.Statement I ($x > y$) is not necessary; the inequality is already true when $x \neq y$.Statement II ($x^2 + y^2 > 2xy$) is equivalent to $(x-y)^2 > 0$, which is also always true for $x \neq y$, but it is not required to prove the result.Thus, neither statement is needed to answer the question.Answer:D – Neither Statement I nor Statement II is required to answer the question.