Sunt binecunoscute urmatoarele doua inegalitati:
1). $\dfrac{x}{y}+\dfrac{y}{x}\ge 2\Leftrightarrow \left(\dfrac{x}{y}+\dfrac{y}{x}\right)^2\ge 4;\,(\forall)x,y>0.$
2).$(x+y)\left(\dfrac{1}{x}+\dfrac{1}{y}\right)\ge 4;\,(\forall)x,y>0.$
Aratati ca: $\boxed{(x+y)\left(\dfrac{1}{x}+\dfrac{1}{y}\right)\le\left(\dfrac{x}{y}+\dfrac{y}{x}\right)^2};\,(\forall)x,y>0.$(OM,Reg.Ural-2015)
OM Reg.URAL(Rusia), 2015
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Re: OM Reg.URAL(Rusia), 2015
Hint : Desfaceți parantezele si pe urma aplicați inegalitatea mediilor.
Liceul Teoretic Cobani
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Re: OM Reg.URAL(Rusia), 2015
Avem:
$\left(\dfrac{x}{y}+\dfrac{y}{x}\right)^2\ge(x+y)\left(\dfrac{1}{x}+\dfrac{1}{y}\right)\Leftrightarrow$$\left(\dfrac{x^2+y^2}{xy}\right)^2\ge\left\dfrac{(x+y)^2}{xy}\right|.x^2y^2\Leftrightarrow$$\left(x^2+y^2\right)^2\ge xy(x+y)^2\Leftrightarrow$
$\Leftrightarrow x^4+2x^2y^2+y^4\ge xy(x+y)^2\Leftrightarrow$$x^4+\underline{2x^2y^2}+y^4\ge x^3y+\underline{2x^2y^2}+xy^3\Leftrightarrow$
$\Leftrightarrow x^4-x^3y-xy^3+y^4\ge 0\Leftrightarrow x^3(x-y)-y^3(x-y)\ge 0$$\Leftrightarrow(x-y)\left(x^3-y^3\right)\ge 0\Leftrightarrow$
$\Leftrightarrow \boxed{(x-y)^2\left(x^2+xy+y^2\right)\ge 0};\,(\forall)x,y>0.$
OBS: O alta inegalitate inrudita cu aceasta a fost data la OMN din ANGLIA-2005,
gasiti aceasta inegalitate pe acest forum, la topicul: http://forum.gil.ro/viewtopic.php?f=19&t=7288
$\left(\dfrac{x}{y}+\dfrac{y}{x}\right)^2\ge(x+y)\left(\dfrac{1}{x}+\dfrac{1}{y}\right)\Leftrightarrow$$\left(\dfrac{x^2+y^2}{xy}\right)^2\ge\left\dfrac{(x+y)^2}{xy}\right|.x^2y^2\Leftrightarrow$$\left(x^2+y^2\right)^2\ge xy(x+y)^2\Leftrightarrow$
$\Leftrightarrow x^4+2x^2y^2+y^4\ge xy(x+y)^2\Leftrightarrow$$x^4+\underline{2x^2y^2}+y^4\ge x^3y+\underline{2x^2y^2}+xy^3\Leftrightarrow$
$\Leftrightarrow x^4-x^3y-xy^3+y^4\ge 0\Leftrightarrow x^3(x-y)-y^3(x-y)\ge 0$$\Leftrightarrow(x-y)\left(x^3-y^3\right)\ge 0\Leftrightarrow$
$\Leftrightarrow \boxed{(x-y)^2\left(x^2+xy+y^2\right)\ge 0};\,(\forall)x,y>0.$
OBS: O alta inegalitate inrudita cu aceasta a fost data la OMN din ANGLIA-2005,
gasiti aceasta inegalitate pe acest forum, la topicul: http://forum.gil.ro/viewtopic.php?f=19&t=7288
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Re: OM Reg.URAL(Rusia), 2015
Vedeti problema propusa PP8 de aici.