Inegalitate cu medii (probabil cunoscuta)

[alex2008]

Daca \( a,b>0 \) sa se arate ca :

\( \frac{2ab}{a+b}+\sqrt{\frac{a^2+b^2}{2}}\ge \sqrt{ab}+\frac{a+b}{2} \)

[maxim bogdan]

Fie \( a,b\in\mathbb{R} \) astfel incat \( ab>0. \) Atunci are loc inegalitatea:

\( \frac{2ab}{a+b}+\sqrt{\frac{a^2+b^2}{2}}\geq\sqrt{ab}+\frac{a+b}{2}. \) (Heinz-Jurgen Seiffert)

Sa introducem substitutiile:

\( a+b=2s \ \wedge \ ab=p^2. \) Din inegalitatea mediilor vom avea ca: \( s\geq p. \)

Inegalitatea din enunt va fi echivalenta cu:

\( \frac{p^2}{s}+\sqrt{2s^2-p^2}\geq s+p \ \Longleftrightarrow \ \sqrt{2s^2-p^2}\geq \frac{s^2+sp-p^2}{s} \ \Longleftrightarrow \ 2s^4-s^2p^2\geq (s^2+sp-p^2)^2 \ \Longleftrightarrow \)

\( \Longleftrightarrow \ 2s^4-s^2p^2\geq s^4+s^2p^2+p^4-2s^2p^2-2p^3s+2s^3p \ \Longleftrightarrow \)

\( \Longleftrightarrow \ s^4-p^4+2p^3s-2s^3p\geq 0 \ \Longleftrightarrow \ (s^2-p^2)(s^2+p^2)-2sp(s^2-p^2)\geq 0 \ \Longleftrightarrow (s^2-p^2)(s-p)^2\geq 0. \)

Ultima inegalitate este adevarata folosind observatia ca \( s\geq p. \) Deci problema e rezolvata.

[Virgil Nicula]

Daca \( a,b>0 \) sa se arate ca \( \frac{2ab}{a+b}+\sqrt{\frac{a^2+b^2}{2}}\ge \sqrt{ab}+\frac{a+b}{2} \) .

Dem. \( \left(\sqrt {ab}+\sqrt {\frac {a^2+b^2}{2}}\right)^2\le 2\cdot \left(ab+\frac {a^2+b^2}{2}\right)=2ab+a^2+b^2=(a+b)^2\ \Longrightarrow\ \underline{\overline{\left\|\ \sqrt {ab}+\sqrt {\frac {a^2+b^2}{2}}\le a+b\ \right\|}}\ (*)\ . \)

Se observa ca \( \frac{2ab}{a+b}+\sqrt{\frac{a^2+b^2}{2}}\ge \sqrt{ab}+\frac{a+b}{2}\ \Longleftrightarrow\ \frac {a+b}{2}-\frac {2ab}{a+b}\le\sqrt {\frac {a^2+b^2}{2}}-\sqrt {ab}\ \Longleftrightarrow \)

\( \left(\frac {a+b}{2}-\frac {2ab}{a+b}\right)\cdot\left(\sqrt {\frac {a^2+b^2}{2}}+\sqrt {ab}\right)\le\frac {a^2+b^2}{2}-ab\ \Longleftrightarrow\ \frac {(a-b)^2}{2(a+b)}\cdot\left(\sqrt {\frac {a^2+b^2}{2}}+\sqrt {ab}\right)\le\frac {(a-b)^2}{2}\ \Longleftrightarrow\ (*)\ . \)

[alex2008]

\( \underline{\overline{\left\|\ \sqrt {ab}+\sqrt {\frac {a^2+b^2}{2}}\le a+b\ \right\|}}\ (*)\ . \)

\( \Leftrightarrow \frac {(a + b)^2}{2}\geq 2\sqrt {\frac {a^2 + b^2}{2}}\sqrt {ab} \) (\( MA-MG \))