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Question Number 202348 by York12 last updated on 24/Dec/23
Let a,b,c ∈R^+ , a+b+c=3 prove the following inequality (((2a−3)^2 )/b)+(((2b−3)^2 )/c)+(((2c−3)^2 )/a)≥((a^2 +b^2 )/(a+b))+((b^2 +c^2 )/(b+c))+((c^2 +a^2 )/(c+a))
$$ \\ $$$$\mathrm{Let}\:{a},{b},{c}\:\:\in\mathbb{R}^{+} \:,\:{a}+{b}+{c}=\mathrm{3}\:\mathrm{prove}\:\mathrm{the}\:\mathrm{following}\:\mathrm{inequality} \\ $$$$\frac{\left(\mathrm{2}{a}−\mathrm{3}\right)^{\mathrm{2}} }{{b}}+\frac{\left(\mathrm{2}{b}−\mathrm{3}\right)^{\mathrm{2}} }{{c}}+\frac{\left(\mathrm{2}{c}−\mathrm{3}\right)^{\mathrm{2}} }{{a}}\geqslant\frac{{a}^{\mathrm{2}} +{b}^{\mathrm{2}} }{{a}+{b}}+\frac{{b}^{\mathrm{2}} +{c}^{\mathrm{2}} }{{b}+{c}}+\frac{{c}^{\mathrm{2}} +{a}^{\mathrm{2}} }{{c}+{a}} \\ $$
Answered by mrdiane last updated on 25/Dec/23
on a 0(a+b+c)^2 =a^2 +b^2 +2(ab+ac+bc comme a,b et >0donc (a+b+c)^2 ≥a^2 +b^2 or b≤a+b donc on a { (((a+b+c)^2 ≥a^2 +b^2 (1))),(((1/b)≥(1/(a+b)) (2))) :} (1)×(2) donne (((a+b+c)^2 )/b)≥ ((a^2 +b^2 )/(a+b))⇔(((a+a−3)^2 )/b)≥((a^2 +b^2 )/(a+b)) pae annalogie on a { (((((2a−3)^2 )/b)≥((a^2 +b^2 )/(a+b)) (3))),(((((2b−3)^2 )/c)≥((b^2 +c^2 )/(b+c)) (4))),(((((2c−3)^2 )/a)≥((c^2 +a^2 )/(a+c)) (5))) :} (3)+(4)+(5) donne (((2a−3)^2 )/b)+(((2b−3))/c)+(((2c−3))/a)≥((a^2 +b^2 )/(a+b))+((b^2 +c^2 )/(b+c))+((c^2 +a^2 )/(c+a))
$$ \\ $$$${on}\:{a}\:\mathrm{0}\left({a}+{b}+{c}\right)^{\mathrm{2}} ={a}^{\mathrm{2}} +{b}^{\mathrm{2}} +\mathrm{2}\left({ab}+{ac}+{bc}\:{comme}\:{a},{b}\:{et}\:>\mathrm{0}{donc}\right. \\ $$$$\left({a}+{b}+{c}\right)^{\mathrm{2}} \geqslant{a}^{\mathrm{2}} +{b}^{\mathrm{2}} \:{or}\:{b}\leqslant{a}+{b} \\ $$$${donc}\:\:{on}\:{a}\:\begin{cases}{\left({a}+{b}+{c}\right)^{\mathrm{2}} \geqslant{a}^{\mathrm{2}} +{b}^{\mathrm{2}} \left(\mathrm{1}\right)}\\{\frac{\mathrm{1}}{{b}}\geqslant\frac{\mathrm{1}}{{a}+{b}}\:\left(\mathrm{2}\right)}\end{cases} \\ $$$$\left(\mathrm{1}\right)×\left(\mathrm{2}\right)\:{donne}\:\frac{\left({a}+{b}+{c}\right)^{\mathrm{2}} }{{b}}\geqslant\:\frac{{a}^{\mathrm{2}} +{b}^{\mathrm{2}} }{{a}+{b}}\Leftrightarrow\frac{\left({a}+{a}−\mathrm{3}\right)^{\mathrm{2}} }{{b}}\geqslant\frac{{a}^{\mathrm{2}} +{b}^{\mathrm{2}} }{{a}+{b}} \\ $$$${pae}\:{annalogie}\:{on}\:{a}\:\begin{cases}{\frac{\left(\mathrm{2}{a}−\mathrm{3}\right)^{\mathrm{2}} }{{b}}\geqslant\frac{{a}^{\mathrm{2}} +{b}^{\mathrm{2}} }{{a}+{b}}\:\left(\mathrm{3}\right)}\\{\frac{\left(\mathrm{2}{b}−\mathrm{3}\right)^{\mathrm{2}} }{{c}}\geqslant\frac{{b}^{\mathrm{2}} +{c}^{\mathrm{2}} }{{b}+{c}}\:\:\left(\mathrm{4}\right)}\\{\frac{\left(\mathrm{2}{c}−\mathrm{3}\right)^{\mathrm{2}} }{{a}}\geqslant\frac{{c}^{\mathrm{2}} +{a}^{\mathrm{2}} }{{a}+{c}}\:\left(\mathrm{5}\right)}\end{cases} \\ $$$$\left(\mathrm{3}\right)+\left(\mathrm{4}\right)+\left(\mathrm{5}\right)\:{donne}\:\frac{\left(\mathrm{2}{a}−\mathrm{3}\right)^{\mathrm{2}} }{{b}}+\frac{\left(\mathrm{2}{b}−\mathrm{3}\right)}{{c}}+\frac{\left(\mathrm{2}{c}−\mathrm{3}\right)}{{a}}\geqslant\frac{{a}^{\mathrm{2}} +{b}^{\mathrm{2}} }{{a}+{b}}+\frac{{b}^{\mathrm{2}} +{c}^{\mathrm{2}} }{{b}+{c}}+\frac{{c}^{\mathrm{2}} +{a}^{\mathrm{2}} }{{c}+{a}} \\ $$$$ \\ $$
Commented by York12 last updated on 25/Dec/23
Merci monsieur, mais je ne comprends pas la première ligne
$$ \\ $$Merci monsieur, mais je ne comprends pas la première ligne
Commented by York12 last updated on 25/Dec/23
(a+b+c)^2 =a^2 +b^2 +c^2 +2(ab+bc+ac) ≠a^2 +b^2 +2(ab+bc+ac) b≥b ⇎ b≥a+b → (because eauality holds when a=0 but a>0 ⇒The equality does not hold) (a+b+c)^2 ≥a^2 +b^2 →(I do not see the equality case) (((a+b+c)^2 )/b)≥ ((a^2 +b^2 )/(a+b))⇔(((a+a−3)^2 )/b)≥((a^2 +b^2 )/(a+b)) →(Why)
$$\left({a}+{b}+{c}\right)^{\mathrm{2}} ={a}^{\mathrm{2}} +{b}^{\mathrm{2}} +{c}^{\mathrm{2}} +\mathrm{2}\left({ab}+{bc}+{ac}\right) \\ $$$$\neq{a}^{\mathrm{2}} +{b}^{\mathrm{2}} +\mathrm{2}\left({ab}+{bc}+{ac}\right) \\ $$$${b}\geqslant{b}\:\nLeftrightarrow\:{b}\geqslant{a}+{b}\:\rightarrow\:\left(\mathrm{because}\:\mathrm{eauality}\:\mathrm{holds}\:\mathrm{when}\:{a}=\mathrm{0}\:{but}\:{a}>\mathrm{0}\:\Rightarrow\mathrm{The}\:\mathrm{equality}\:\mathrm{does}\:\mathrm{not}\:\mathrm{hold}\right) \\ $$$$\left({a}+{b}+{c}\right)^{\mathrm{2}} \geqslant{a}^{\mathrm{2}} +{b}^{\mathrm{2}} \:\rightarrow\left(\mathrm{I}\:\mathrm{do}\:\mathrm{not}\:\mathrm{see}\:\mathrm{the}\:\mathrm{equality}\:\mathrm{case}\right) \\ $$$$\frac{\left({a}+{b}+{c}\right)^{\mathrm{2}} }{{b}}\geqslant\:\frac{{a}^{\mathrm{2}} +{b}^{\mathrm{2}} }{{a}+{b}}\Leftrightarrow\frac{\left({a}+{a}−\mathrm{3}\right)^{\mathrm{2}} }{{b}}\geqslant\frac{{a}^{\mathrm{2}} +{b}^{\mathrm{2}} }{{a}+{b}}\:\rightarrow\left(\mathrm{Why}\right) \\ $$

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