Aplicatie(Pb 3, Concursul "Marian Tarina", 2004)

[Claudiu Mindrila]

Fie \( a,b,c,d>0 \) astfel incat \( a+b+c+d=16 \). Sa se demonstreze inegalitatea:

\(
\left( {a + \frac{1}
{c}} \right)^2 + \left( {c + \frac{1}
{a}} \right)^2 + \left( {b + \frac{1}
{d}} \right)^2 + \left( {d + \frac{1}
{b}} \right)^2 \geq \frac{{289}}
{4}

\)
Mircea Lascu, Marian Tetiva, Concursul "Marian Tarina", 2004

[Omer Cerrahoglu]

Inegalitatea este echivalenta cu \( a^2+b^2+c^2+d^2+2(\frac{a^2}{c^2}+\frac{c^2}{a^2}+\frac{b^2}{d^2}+\frac{d^2}{b^2})+\frac{1}{a^2}+\frac{1}{b^2}+\frac{1}{c^2}+\frac{1}{d^2}\geq \frac{289}{4}(*) \).
Din inegalitatea mediilor avem ca \( a^2+b^2+c^2+d^2\geq 64 (1) \)
Dar avem ca \( (a+b+c+d)(\frac{1}{a}+\frac{1}{b}+\frac{1}{c}+\frac{1}{d}) \geq 16 \), deci folosind ipoteza avem ca \( \frac{1}{a}+\frac{1}{b}+\frac{1}{c}+\frac{1}{d} \geq 1 \). Din inegalitatea mediilor avem ca \( \frac{1}{a^2}+\frac{1}{b^2}+\frac{1}{c^2}+\frac{1}{d^2}\geq \frac{1}{4}(2) \).
Avem ca \( 2(\frac{a^2}{c^2}+\frac{c^2}{a^2}+\frac{b^2}{d^2}+\frac{d^2}{b^2})\geq 8(3) \).
Insumand (1), (2) si (3) obtinem (*), deci inegaliatatea este demonstrata

[Claudiu Mindrila]

Draguta solutie. Uite si solutia mea:
Din \( MA-MP \) avem \( \frac{x+y+z+t}{4} \leq \sqrt{\frac{x^2+y^2+z^2+t^2}{4}}\Rightarrow 4(x^2+y^2+z^2+t^2) \geq (x+y+z+t)^2 \), unde \( x,y,z,t \geq 0 \).
Revenind la problema, avem ca:
\( LHS \geq \frac{(a+b+c+d+\frac{1}{a}+\frac{1}{b}+\frac{1}{c}+\frac{1}{d})^2}{4}\geq^{CBS}\frac{(16+\frac{16}{a+b+c+d})^2}{4}=\frac{17^2}{4}=\frac{289}{4} \)