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Fie \( \triangle ABC \) cu \( A=90^{\circ}\ . \) Sa se arate ca \( \frac {a^2}{bc}+\frac {b+c}{a}\ \ge\ 2+\sqrt 2 \) (Crux Mathematicorum).

Generalizare proprie. Fie \( \triangle ABC \). Sa se arate ca daca

\( a=\max\ \{a,b,c\}\ \Longrightarrow\ \frac {b^2+c^2}{bc}+\frac {b+c}{a}\ \ge\ 2+\frac {1}{\sin\frac A2}\cdot\sqrt {1+\left(\frac {b-c}{2a}\right)^2}\ . \)

Virgil Nicula wrote:Fie \( \triangle ABC \) cu \( A=90^{\circ}\ . \) Sa se arate ca \( \frac {a^2}{bc}+\frac {b+c}{a}\ \ge\ 2+\sqrt 2 \) (Crux Mathematicorum).

Inegalitatea este echivalenta cu \( (\frac{b^2+c^2}{bc}+\frac{b+c}{\sqrt{b^2+c^2}})^2\ge 6+4\sqrt{2} \) (*)

Insa \( \frac{(b^2+c^2)^2}{b^2c^2}+\frac{(b+c)^2}{b^2+c^2}\ge3\sqrt[3]{\frac{(b^2+c^2)^4}{4b^4c^4}\cdot\frac{(b+c)^2}{b^2+c^2}}\ge 6 \) si

\( 2\cdot\frac{b^2+c^2}{bc}\cdot\frac{(b+c)^2}{b^2+c^2}\ge 4\sqrt{2} \) si atunci (*) este adevarata.