Итак, нужно доказать, что:
![$\[a + b + c + d \ge 3(\frac{1}{a} + \frac{1}{b} + \frac{1}{c} + \frac{1}{d})\]$ $\[a + b + c + d \ge 3(\frac{1}{a} + \frac{1}{b} + \frac{1}{c} + \frac{1}{d})\]$](https://dxdy-03.korotkov.co.uk/f/6/a/b/6abf2361a26c4633ea0b6aff107e66ad82.png)
Из условия нетрудно получить, что:
![$\[a = \sqrt {\frac{{\frac{1}{{1 + {b^2}}} + \frac{1}{{1 + {c^2}}} + \frac{1}{{1 + {d^2}}}}}{{\frac{1}{{1 + {a^2}}}}}} \]$ $\[a = \sqrt {\frac{{\frac{1}{{1 + {b^2}}} + \frac{1}{{1 + {c^2}}} + \frac{1}{{1 + {d^2}}}}}{{\frac{1}{{1 + {a^2}}}}}} \]$](https://dxdy-01.korotkov.co.uk/f/0/2/f/02f5845fd41c148189dfce9db7bde98382.png)
,
![$\[b = \sqrt {\frac{{\frac{1}{{1 + {a^2}}} + \frac{1}{{1 + {c^2}}} + \frac{1}{{1 + {d^2}}}}}{{\frac{1}{{1 + {b^2}}}}}} \]$ $\[b = \sqrt {\frac{{\frac{1}{{1 + {a^2}}} + \frac{1}{{1 + {c^2}}} + \frac{1}{{1 + {d^2}}}}}{{\frac{1}{{1 + {b^2}}}}}} \]$](https://dxdy-04.korotkov.co.uk/f/3/3/1/33126c78b248c74bc692b660762e5f9982.png)
,
![$\[c = \sqrt {\frac{{\frac{1}{{1 + {a^2}}} + \frac{1}{{1 + {b^2}}} + \frac{1}{{1 + {d^2}}}}}{{\frac{1}{{1 + {c^2}}}}}} \]$ $\[c = \sqrt {\frac{{\frac{1}{{1 + {a^2}}} + \frac{1}{{1 + {b^2}}} + \frac{1}{{1 + {d^2}}}}}{{\frac{1}{{1 + {c^2}}}}}} \]$](https://dxdy-04.korotkov.co.uk/f/7/6/8/7689771c3f8035f80436feeb5e1c76ae82.png)
и
![$\[d = \sqrt {\frac{{\frac{1}{{1 + {a^2}}} + \frac{1}{{1 + {b^2}}} + \frac{1}{{1 + {c^2}}}}}{{\frac{1}{{1 + {d^2}}}}}} \]$ $\[d = \sqrt {\frac{{\frac{1}{{1 + {a^2}}} + \frac{1}{{1 + {b^2}}} + \frac{1}{{1 + {c^2}}}}}{{\frac{1}{{1 + {d^2}}}}}} \]$](https://dxdy-02.korotkov.co.uk/f/9/b/2/9b22ccf7bba8dcf823284ad10227248882.png)
.
И наше неравенство запишется так:
![$\[\sqrt {\frac{{\frac{1}{{1 + {b^2}}} + \frac{1}{{1 + {c^2}}} + \frac{1}{{1 + {d^2}}}}}{{\frac{1}{{1 + {a^2}}}}}} + \sqrt {\frac{{\frac{1}{{1 + {a^2}}} + \frac{1}{{1 + {c^2}}} + \frac{1}{{1 + {d^2}}}}}{{\frac{1}{{1 + {b^2}}}}}} + \sqrt {\frac{{\frac{1}{{1 + {a^2}}} + \frac{1}{{1 + {b^2}}} + \frac{1}{{1 + {d^2}}}}}{{\frac{1}{{1 + {c^2}}}}}} + \sqrt {\frac{{\frac{1}{{1 + {a^2}}} + \frac{1}{{1 + {b^2}}} + \frac{1}{{1 + {c^2}}}}}{{\frac{1}{{1 + {d^2}}}}}} \ge 3\left( {\sqrt {\frac{{\frac{1}{{1 + {a^2}}}}}{{\frac{1}{{1 + {b^2}}} + \frac{1}{{1 + {c^2}}} + \frac{1}{{1 + {d^2}}}}}} + \sqrt {\frac{{\frac{1}{{1 + {b^2}}}}}{{\frac{1}{{1 + {a^2}}} + \frac{1}{{1 + {c^2}}} + \frac{1}{{1 + {d^2}}}}}} + \sqrt {\frac{{\frac{1}{{1 + {c^2}}}}}{{\frac{1}{{1 + {a^2}}} + \frac{1}{{1 + {b^2}}} + \frac{1}{{1 + {d^2}}}}}} + \sqrt {\frac{{\frac{1}{{1 + {d^2}}}}}{{\frac{1}{{1 + {a^2}}} + \frac{1}{{1 + {b^2}}} + \frac{1}{{1 + {c^2}}}}}} } \right)\]$ $\[\sqrt {\frac{{\frac{1}{{1 + {b^2}}} + \frac{1}{{1 + {c^2}}} + \frac{1}{{1 + {d^2}}}}}{{\frac{1}{{1 + {a^2}}}}}} + \sqrt {\frac{{\frac{1}{{1 + {a^2}}} + \frac{1}{{1 + {c^2}}} + \frac{1}{{1 + {d^2}}}}}{{\frac{1}{{1 + {b^2}}}}}} + \sqrt {\frac{{\frac{1}{{1 + {a^2}}} + \frac{1}{{1 + {b^2}}} + \frac{1}{{1 + {d^2}}}}}{{\frac{1}{{1 + {c^2}}}}}} + \sqrt {\frac{{\frac{1}{{1 + {a^2}}} + \frac{1}{{1 + {b^2}}} + \frac{1}{{1 + {c^2}}}}}{{\frac{1}{{1 + {d^2}}}}}} \ge 3\left( {\sqrt {\frac{{\frac{1}{{1 + {a^2}}}}}{{\frac{1}{{1 + {b^2}}} + \frac{1}{{1 + {c^2}}} + \frac{1}{{1 + {d^2}}}}}} + \sqrt {\frac{{\frac{1}{{1 + {b^2}}}}}{{\frac{1}{{1 + {a^2}}} + \frac{1}{{1 + {c^2}}} + \frac{1}{{1 + {d^2}}}}}} + \sqrt {\frac{{\frac{1}{{1 + {c^2}}}}}{{\frac{1}{{1 + {a^2}}} + \frac{1}{{1 + {b^2}}} + \frac{1}{{1 + {d^2}}}}}} + \sqrt {\frac{{\frac{1}{{1 + {d^2}}}}}{{\frac{1}{{1 + {a^2}}} + \frac{1}{{1 + {b^2}}} + \frac{1}{{1 + {c^2}}}}}} } \right)\]$](https://dxdy-01.korotkov.co.uk/f/8/8/5/885e125d8cd6e313d4e018bab38f65a582.png)
Применяя Коши-Шварца имеем
![$\[\begin{array}{l}
\sqrt {\frac{{\frac{1}{{1 + {b^2}}} + \frac{1}{{1 + {c^2}}} + \frac{1}{{1 + {d^2}}}}}{{\frac{1}{{1 + {a^2}}}}}} \ge \frac{{\sqrt {\frac{1}{{1 + {b^2}}}} + \sqrt {\frac{1}{{1 + {c^2}}}} + \sqrt {\frac{1}{{1 + {d^2}}}} }}{{\sqrt 3 \sqrt {\frac{1}{{1 + {a^2}}}} }} \\
\sqrt {\frac{{\frac{1}{{1 + {a^2}}} + \frac{1}{{1 + {c^2}}} + \frac{1}{{1 + {d^2}}}}}{{\frac{1}{{1 + {b^2}}}}}} \ge \frac{{\sqrt {\frac{1}{{1 + {a^2}}}} + \sqrt {\frac{1}{{1 + {c^2}}}} + \sqrt {\frac{1}{{1 + {d^2}}}} }}{{\sqrt 3 \sqrt {\frac{1}{{1 + {b^2}}}} }} \\
\sqrt {\frac{{\frac{1}{{1 + {a^2}}} + \frac{1}{{1 + {b^2}}} + \frac{1}{{1 + {d^2}}}}}{{\frac{1}{{1 + {c^2}}}}}} \ge \frac{{\sqrt {\frac{1}{{1 + {a^2}}}} + \sqrt {\frac{1}{{1 + {b^2}}}} + \sqrt {\frac{1}{{1 + {d^2}}}} }}{{\sqrt 3 \sqrt {\frac{1}{{1 + {c^2}}}} }} \\
\sqrt {\frac{{\frac{1}{{1 + {a^2}}} + \frac{1}{{1 + {b^2}}} + \frac{1}{{1 + {c^2}}}}}{{\frac{1}{{1 + {d^2}}}}}} \ge \frac{{\sqrt {\frac{1}{{1 + {a^2}}}} + \sqrt {\frac{1}{{1 + {b^2}}}} + \sqrt {\frac{1}{{1 + {c^2}}}} }}{{\sqrt 3 \sqrt {\frac{1}{{1 + {d^2}}}} }} \\
\end{array}\]$ $\[\begin{array}{l}
\sqrt {\frac{{\frac{1}{{1 + {b^2}}} + \frac{1}{{1 + {c^2}}} + \frac{1}{{1 + {d^2}}}}}{{\frac{1}{{1 + {a^2}}}}}} \ge \frac{{\sqrt {\frac{1}{{1 + {b^2}}}} + \sqrt {\frac{1}{{1 + {c^2}}}} + \sqrt {\frac{1}{{1 + {d^2}}}} }}{{\sqrt 3 \sqrt {\frac{1}{{1 + {a^2}}}} }} \\
\sqrt {\frac{{\frac{1}{{1 + {a^2}}} + \frac{1}{{1 + {c^2}}} + \frac{1}{{1 + {d^2}}}}}{{\frac{1}{{1 + {b^2}}}}}} \ge \frac{{\sqrt {\frac{1}{{1 + {a^2}}}} + \sqrt {\frac{1}{{1 + {c^2}}}} + \sqrt {\frac{1}{{1 + {d^2}}}} }}{{\sqrt 3 \sqrt {\frac{1}{{1 + {b^2}}}} }} \\
\sqrt {\frac{{\frac{1}{{1 + {a^2}}} + \frac{1}{{1 + {b^2}}} + \frac{1}{{1 + {d^2}}}}}{{\frac{1}{{1 + {c^2}}}}}} \ge \frac{{\sqrt {\frac{1}{{1 + {a^2}}}} + \sqrt {\frac{1}{{1 + {b^2}}}} + \sqrt {\frac{1}{{1 + {d^2}}}} }}{{\sqrt 3 \sqrt {\frac{1}{{1 + {c^2}}}} }} \\
\sqrt {\frac{{\frac{1}{{1 + {a^2}}} + \frac{1}{{1 + {b^2}}} + \frac{1}{{1 + {c^2}}}}}{{\frac{1}{{1 + {d^2}}}}}} \ge \frac{{\sqrt {\frac{1}{{1 + {a^2}}}} + \sqrt {\frac{1}{{1 + {b^2}}}} + \sqrt {\frac{1}{{1 + {c^2}}}} }}{{\sqrt 3 \sqrt {\frac{1}{{1 + {d^2}}}} }} \\
\end{array}\]$](https://dxdy-02.korotkov.co.uk/f/9/d/1/9d122acb81387dc6b8da17428854fddb82.png)
А теперь применив, сперва AM-GM, затем слегка попереставляем слагаемые и затем снова применив Коши-Шварца (для сворачивания), получаем окончательно:
![$\[\begin{array}{l}
LHS \ge \frac{{\sqrt {\frac{1}{{1 + {a^2}}}} }}{{\sqrt 3 }}\left( {\frac{1}{{\sqrt {\frac{1}{{1 + {b^2}}}} }} + \frac{1}{{\sqrt {\frac{1}{{1 + {c^2}}}} }} + \frac{1}{{\sqrt {\frac{1}{{1 + {d^2}}}} }}} \right) + \frac{{\sqrt {\frac{1}{{1 + {b^2}}}} }}{{\sqrt 3 }}\left( {\frac{1}{{\sqrt {\frac{1}{{1 + {a^2}}}} }} + \frac{1}{{\sqrt {\frac{1}{{1 + {c^2}}}} }} + \frac{1}{{\sqrt {\frac{1}{{1 + {d^2}}}} }}} \right) + \frac{{\sqrt {\frac{1}{{1 + {c^2}}}} }}{{\sqrt 3 }}\left( {\frac{1}{{\sqrt {\frac{1}{{1 + {a^2}}}} }} + \frac{1}{{\sqrt {\frac{1}{{1 + {v^2}}}} }} + \frac{1}{{\sqrt {\frac{1}{{1 + {d^2}}}} }}} \right) + \frac{{\sqrt {\frac{1}{{1 + {d^2}}}} }}{{\sqrt 3 }}\left( {\frac{1}{{\sqrt {\frac{1}{{1 + {a^2}}}} }} + \frac{1}{{\sqrt {\frac{1}{{1 + {b^2}}}} }} + \frac{1}{{\sqrt {\frac{1}{{1 + {c^2}}}} }}} \right) \\
\ge \frac{{3\sqrt 3 \sqrt {\frac{1}{{1 + {a^2}}}} }}{{\sqrt {\frac{1}{{1 + {b^2}}}} + \sqrt {\frac{1}{{1 + {c^2}}}} + \sqrt {\frac{1}{{1 + {d^2}}}} }} + \frac{{3\sqrt 3 \sqrt {\frac{1}{{1 + {b^2}}}} }}{{\sqrt {\frac{1}{{1 + {a^2}}}} + \sqrt {\frac{1}{{1 + {c^2}}}} + \sqrt {\frac{1}{{1 + {d^2}}}} }} + \frac{{3\sqrt 3 \sqrt {\frac{1}{{1 + {c^2}}}} }}{{\sqrt {\frac{1}{{1 + {a^2}}}} + \sqrt {\frac{1}{{1 + {b^2}}}} + \sqrt {\frac{1}{{1 + {d^2}}}} }} + \frac{{3\sqrt 3 \sqrt {\frac{1}{{1 + {d^2}}}} }}{{\sqrt {\frac{1}{{1 + {a^2}}}} + \sqrt {\frac{1}{{1 + {b^2}}}} + \sqrt {\frac{1}{{1 + {c^2}}}} }} \\
\ge \left( {\sqrt {\frac{{\frac{1}{{1 + {a^2}}}}}{{\frac{1}{{1 + {b^2}}} + \frac{1}{{1 + {c^2}}} + \frac{1}{{1 + {d^2}}}}}} + \sqrt {\frac{{\frac{1}{{1 + {b^2}}}}}{{\frac{1}{{1 + {a^2}}} + \frac{1}{{1 + {c^2}}} + \frac{1}{{1 + {d^2}}}}}} + \sqrt {\frac{{\frac{1}{{1 + {c^2}}}}}{{\frac{1}{{1 + {a^2}}} + \frac{1}{{1 + {b^2}}} + \frac{1}{{1 + {d^2}}}}}} + \sqrt {\frac{{\frac{1}{{1 + {d^2}}}}}{{\frac{1}{{1 + {a^2}}} + \frac{1}{{1 + {b^2}}} + \frac{1}{{1 + {c^2}}}}}} } \right) \\
\end{array}\]$ $\[\begin{array}{l}
LHS \ge \frac{{\sqrt {\frac{1}{{1 + {a^2}}}} }}{{\sqrt 3 }}\left( {\frac{1}{{\sqrt {\frac{1}{{1 + {b^2}}}} }} + \frac{1}{{\sqrt {\frac{1}{{1 + {c^2}}}} }} + \frac{1}{{\sqrt {\frac{1}{{1 + {d^2}}}} }}} \right) + \frac{{\sqrt {\frac{1}{{1 + {b^2}}}} }}{{\sqrt 3 }}\left( {\frac{1}{{\sqrt {\frac{1}{{1 + {a^2}}}} }} + \frac{1}{{\sqrt {\frac{1}{{1 + {c^2}}}} }} + \frac{1}{{\sqrt {\frac{1}{{1 + {d^2}}}} }}} \right) + \frac{{\sqrt {\frac{1}{{1 + {c^2}}}} }}{{\sqrt 3 }}\left( {\frac{1}{{\sqrt {\frac{1}{{1 + {a^2}}}} }} + \frac{1}{{\sqrt {\frac{1}{{1 + {v^2}}}} }} + \frac{1}{{\sqrt {\frac{1}{{1 + {d^2}}}} }}} \right) + \frac{{\sqrt {\frac{1}{{1 + {d^2}}}} }}{{\sqrt 3 }}\left( {\frac{1}{{\sqrt {\frac{1}{{1 + {a^2}}}} }} + \frac{1}{{\sqrt {\frac{1}{{1 + {b^2}}}} }} + \frac{1}{{\sqrt {\frac{1}{{1 + {c^2}}}} }}} \right) \\
\ge \frac{{3\sqrt 3 \sqrt {\frac{1}{{1 + {a^2}}}} }}{{\sqrt {\frac{1}{{1 + {b^2}}}} + \sqrt {\frac{1}{{1 + {c^2}}}} + \sqrt {\frac{1}{{1 + {d^2}}}} }} + \frac{{3\sqrt 3 \sqrt {\frac{1}{{1 + {b^2}}}} }}{{\sqrt {\frac{1}{{1 + {a^2}}}} + \sqrt {\frac{1}{{1 + {c^2}}}} + \sqrt {\frac{1}{{1 + {d^2}}}} }} + \frac{{3\sqrt 3 \sqrt {\frac{1}{{1 + {c^2}}}} }}{{\sqrt {\frac{1}{{1 + {a^2}}}} + \sqrt {\frac{1}{{1 + {b^2}}}} + \sqrt {\frac{1}{{1 + {d^2}}}} }} + \frac{{3\sqrt 3 \sqrt {\frac{1}{{1 + {d^2}}}} }}{{\sqrt {\frac{1}{{1 + {a^2}}}} + \sqrt {\frac{1}{{1 + {b^2}}}} + \sqrt {\frac{1}{{1 + {c^2}}}} }} \\
\ge \left( {\sqrt {\frac{{\frac{1}{{1 + {a^2}}}}}{{\frac{1}{{1 + {b^2}}} + \frac{1}{{1 + {c^2}}} + \frac{1}{{1 + {d^2}}}}}} + \sqrt {\frac{{\frac{1}{{1 + {b^2}}}}}{{\frac{1}{{1 + {a^2}}} + \frac{1}{{1 + {c^2}}} + \frac{1}{{1 + {d^2}}}}}} + \sqrt {\frac{{\frac{1}{{1 + {c^2}}}}}{{\frac{1}{{1 + {a^2}}} + \frac{1}{{1 + {b^2}}} + \frac{1}{{1 + {d^2}}}}}} + \sqrt {\frac{{\frac{1}{{1 + {d^2}}}}}{{\frac{1}{{1 + {a^2}}} + \frac{1}{{1 + {b^2}}} + \frac{1}{{1 + {c^2}}}}}} } \right) \\
\end{array}\]$](https://dxdy-04.korotkov.co.uk/f/7/b/1/7b13751257360c9bb33b8c6b0422b11782.png)
На этом и завершаем доказательство.
P.S. Решение уродливое. Красивое решение могло бы быть при использовании неравенства Чебышева, но мне не удалось обосновать его применение.
Точнее удалось, но только в том случае, если все 4 искомых числа больше 1.