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Posted: **Wed Nov 19, 2008 10:00 am**

Fie doua numere pozitive \( a \), \( b \). Se arata usor ca \( (a+b)^4\le \left(a^2+3b^2\right)\left(b^2+3a^2\right). \)

Sa se arate o "intarire" a acesteia : \( (a+b)^4\le \overline {\underline {\left\|\ (a+b)^3\sqrt {2\left(a^2+b^2\right)}\le \left(a^2+3b^2\right)\left(b^2+3a^2\right)\ \right\|}}\ . \)

Posted: **Wed Nov 19, 2008 10:56 pm**

RHS=\( (a^2+b^2+b^2+b^2)(a^2+a^2+a^2+b^2)\ge (aa+ba+ba+bb)^2=(a+b)^4 \)

Posted: **Sun Sep 20, 2009 11:18 am**

\( ((a^2+3b^2)(3a^2+b^2))^2-2(a+b)^6(a^2+b^2)=2((a^3-b^3)(a-b))^2+2ab(a^4+b^4)(a-b)^2+5(a^6+b^6)(a-b)^2+25a^2b^2(a^2+b^2)(a-b)^2 \)

Posted: **Sun Sep 20, 2009 7:12 pm**

Virgil Nicula wrote: \( \overline {\underline {\left\|\ (a+b)^3\sqrt {2\left(a^2+b^2\right)}\le \left(a^2+3b^2\right)\left(b^2+3a^2\right)\ \right\|}}. \)

Notam \( \frac{a}{b}=x>0 \) si inegalitatea este echivalenta cu

\( (x^2+3)^2(1+3x^2)^2-2(x+1)^6(x^1+1)\ge 0 \)

sau

\( (x-1)^2(7x^6+2x^5+25x^4-4x^3+25x^2+2x+7)\ge 0 \)

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O inegalitate in doua variabile pozitive.

O inegalitate in doua variabile pozitive.

Re: O inegalitate in doua variabile pozitive.