Assume n ≥ 3 , c ≥ 1 , d ≥ 1 are natural numbers such that c ² + d ² − ( n − 1 ) c d &lt; 0. Show that ( n ³ − n + 1 ) c ² + ( n + 1 ) d ² − ( n ² + n − 1 ) c d &gt; 1

Question

Assume   n  ≥  3  ,  c  ≥  1  ,  d  ≥  1 are natural numbers such that   c      ²    +  d      ²    −  (  n  −  1  )  c  d  &amp;lt;  0. Show that   (  n      ³    −  n  +  1  )  c      ²    +  (  n  +  1  )  d      ²    −  (  n      ²    +  n  −  1  )  c  d  &amp;gt;  1

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Let   c  ,  d be positive real numbers and   n  &amp;gt;  1 (esp.,       n    3    −  n  +  1  &amp;gt;  0). By the arithmetic-geometric inequality  (      n    3    −  n  +  1  )      c    2    +  (  n  +  1  )      d    2    ≥  2  ⋅      (          n      3        −    n    +    1    )    (    n    +    1    )    ⋅  c  d  .One checks by multiplying out that  4  (      n    3    −  n  +  1  )  (  n  +  1  )  =  (      n    2    +  n  −  1      )    2    +  3  +  3      n    2    (      n    2    −  1  )  +  2  n  (      n    2    +  1  )  ,hence   2  ⋅      (          n      3        −    n    +    1    )    (    n    +    1    )    &amp;gt;      n    2    +  n  −  1 and finally  (      n    3    −  n  +  1  )      c    2    +  (  n  +  1  )      d    2    &amp;gt;  (      n    2    +  n  −  1  )  c  d  .

Assume n>=3, c>=1, d>=1 are natural numbers such that c^2+d^2−(n−1)cd<0. Show that (n^3−n+1)c^2+(n+1)d^2−(n^2+n−1)cd>1.

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