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Question Number 205423 by hardmath last updated on 20/Mar/24
If a , b ∈ R Then: a^2 + b^2 ≥ ab + (√((a^4 + b^4 )/2))
$$\mathrm{If} \\ $$$$\mathrm{a}\:,\:\mathrm{b}\:\in\:\mathbb{R} \\ $$$$\mathrm{Then}: \\ $$$$\mathrm{a}^{\mathrm{2}} \:+\:\mathrm{b}^{\mathrm{2}} \:\geqslant\:\mathrm{ab}\:+\:\sqrt{\frac{\mathrm{a}^{\mathrm{4}} \:+\:\mathrm{b}^{\mathrm{4}} }{\mathrm{2}}} \\ $$
Answered by Berbere last updated on 21/Mar/24
⇔(a^2 +b^2 −ab)^2 ≥((a^4 +b^4 )/2) ⇔((a^4 +b^4 )/2)+3a^2 b^2 −2ab(a^2 +b^2 )≥0 a^4 +b^4 =(a^2 +b^2 )^2 −2a^2 b^2 ⇔(((a^2 +b^2 )^2 −2a^2 b^2 )/2)+3a^2 b^2 −2ab(a^2 +b^2 )≥0 ⇔(a^2 +b^2 )^2 +(2ab)^2 −2(2ab)(a^2 +b^2 )≥0 ⇔(a^2 +b^2 −2ab)^2 ≥0True
$$\Leftrightarrow\left({a}^{\mathrm{2}} +{b}^{\mathrm{2}} −{ab}\right)^{\mathrm{2}} \geqslant\frac{{a}^{\mathrm{4}} +{b}^{\mathrm{4}} }{\mathrm{2}} \\ $$$$\Leftrightarrow\frac{{a}^{\mathrm{4}} +{b}^{\mathrm{4}} }{\mathrm{2}}+\mathrm{3}{a}^{\mathrm{2}} {b}^{\mathrm{2}} −\mathrm{2}{ab}\left({a}^{\mathrm{2}} +{b}^{\mathrm{2}} \right)\geqslant\mathrm{0} \\ $$$${a}^{\mathrm{4}} +{b}^{\mathrm{4}} =\left({a}^{\mathrm{2}} +{b}^{\mathrm{2}} \right)^{\mathrm{2}} −\mathrm{2}{a}^{\mathrm{2}} {b}^{\mathrm{2}} \\ $$$$\Leftrightarrow\frac{\left({a}^{\mathrm{2}} +{b}^{\mathrm{2}} \right)^{\mathrm{2}} −\mathrm{2}{a}^{\mathrm{2}} {b}^{\mathrm{2}} }{\mathrm{2}}+\mathrm{3}{a}^{\mathrm{2}} {b}^{\mathrm{2}} −\mathrm{2}{ab}\left({a}^{\mathrm{2}} +{b}^{\mathrm{2}} \right)\geqslant\mathrm{0} \\ $$$$\Leftrightarrow\left({a}^{\mathrm{2}} +{b}^{\mathrm{2}} \right)^{\mathrm{2}} +\left(\mathrm{2}{ab}\right)^{\mathrm{2}} −\mathrm{2}\left(\mathrm{2}{ab}\right)\left({a}^{\mathrm{2}} +{b}^{\mathrm{2}} \right)\geqslant\mathrm{0} \\ $$$$\Leftrightarrow\left({a}^{\mathrm{2}} +{b}^{\mathrm{2}} −\mathrm{2}{ab}\right)^{\mathrm{2}} \geqslant\mathrm{0}{True} \\ $$$$ \\ $$$$ \\ $$
Commented by Skabetix last updated on 22/Mar/24
comment passer de (a^2 +b^2 −ab)^2 ≥((a^4 +b^4 )/2) a ((a^4 +b^4 )/2)+3a^2 b^2 −2ab(a^2 +b^2 )≥0?
$$\mathrm{comment}\:\mathrm{passer}\:\mathrm{de}\: \\ $$$$\left(\mathrm{a}^{\mathrm{2}} +\mathrm{b}^{\mathrm{2}} −\mathrm{ab}\right)^{\mathrm{2}} \:\geqslant\frac{\mathrm{a}^{\mathrm{4}} +\mathrm{b}^{\mathrm{4}} }{\mathrm{2}} \\ $$$$\mathrm{a}\:\:\:\:\:\frac{\mathrm{a}^{\mathrm{4}} +\mathrm{b}^{\mathrm{4}} }{\mathrm{2}}+\mathrm{3a}^{\mathrm{2}} \mathrm{b}^{\mathrm{2}} −\mathrm{2ab}\left(\mathrm{a}^{\mathrm{2}} +\mathrm{b}^{\mathrm{2}} \right)\geqslant\mathrm{0}? \\ $$
Commented by Berbere last updated on 24/Mar/24
squar a^4 +b^4 +2a^2 b^2 +a^2 b^2 −2ab(a^2 +b^2 )≥((a^4 +b^4 )/2) ⇒a^4 +b^4 −((a^4 +b^4 )/2)+3a^2 b^2 −2ab(a^2 +b^2 )≥0 ((a^4 +b^4 )/2)+3a^2 b^2 −2ab(a^2 +b^2 )≥0
$${squar} \\ $$$${a}^{\mathrm{4}} +{b}^{\mathrm{4}} +\mathrm{2}{a}^{\mathrm{2}} {b}^{\mathrm{2}} +{a}^{\mathrm{2}} {b}^{\mathrm{2}} −\mathrm{2}{ab}\left({a}^{\mathrm{2}} +{b}^{\mathrm{2}} \right)\geqslant\frac{{a}^{\mathrm{4}} +{b}^{\mathrm{4}} }{\mathrm{2}} \\ $$$$\Rightarrow{a}^{\mathrm{4}} +{b}^{\mathrm{4}} −\frac{{a}^{\mathrm{4}} +{b}^{\mathrm{4}} }{\mathrm{2}}+\mathrm{3}{a}^{\mathrm{2}} {b}^{\mathrm{2}} −\mathrm{2}{ab}\left({a}^{\mathrm{2}} +{b}^{\mathrm{2}} \right)\geqslant\mathrm{0} \\ $$$$\frac{{a}^{\mathrm{4}} +{b}^{\mathrm{4}} }{\mathrm{2}}+\mathrm{3}{a}^{\mathrm{2}} {b}^{\mathrm{2}} −\mathrm{2}{ab}\left({a}^{\mathrm{2}} +{b}^{\mathrm{2}} \right)\geqslant\mathrm{0} \\ $$

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