Question-226983

Question Number 226983 by hardmath last updated on 24/Dec/25

Answered by MrAjder last updated on 01/Jan/26

S={(a,b,c)∈R_+ ^3 :a+b+c=3},F(a,b,c)=Σ_(cyc) (a/(b^2 +c)) L(a,b,c,λ)=F(a,b,c)+λ(a+b+c−3) (∂L/∂a)=(1/(b^2 +c))−(b/((c^2 +a^2 )^2 ))−((2ac)/((a^2 +b)^2 ))+λ=0 (∂L/∂b)=(1/(c^2 +a))−(c/((a^2 +b)^2 ))−((2ba)/((b^2 +c^2 )))+λ=0 (∂L/∂c)=(1/(a^2 +b))−(a/((b^2 +c)^2 ))−((2ab)/((c^2 +a)^2 ))+λ=0 a=b=c,a+b+c=3⇒a=b=c=1 λ=−(1/4) H_L (1,1,1)= (((∂^2 F/∂a^2 ),(∂^2 F/(∂a∂b)),(∂^2 F/(∂a∂c))),((∂^2 F/(∂b∂a)),(∂^2 F/∂b^2 ),(∂^2 F/(∂b∂c))),((∂^2 F/(∂c∂b)),(∂^2 F/(∂c∂b)),(∂^2 F/∂c^2 )) )_((1,1,1)) (∂^2 F/∂a^2 )∣_((1,1,1)) =(3/4),(∂^2 F/(∂a∂b))∣_((1,1,1)) =(1/4) H= (((3/4),(−(1/4)),(−(1/4))),((−(1/4)),(3/4),(−(1/4))),((−(1/4)),(−(1/4)),(3/4)) ) det ((((3/4)−μ),(−(1/4)),(−(1/4))),((−(1/4)),((3/4)−μ),(−(1/4))),((−(1/4)),(−(1/4)),((3/4)−μ)) )=0⇒(μ−(1/4))(μ−1)^2 =0 σ(H)={(1/4)1,1}⊂R^+ ∴H≻0(1,1,1)Is a strictly local minimum point min_(S) F=F(1,1,1)=(1/(1^2 +1))+(1/(1^2 +1))+(1/(1^2 +1))=(3/2) ∴∀(a,b,c)∈S,Σ_(cyc) (a/(b^2 −c))≥(3/2)

$$ \\ $$$${S}=\left\{\left({a},{b},{c}\right)\in\mathbb{R}_{+} ^{\mathrm{3}} :{a}+{b}+{c}=\mathrm{3}\right\},{F}\left({a},{b},{c}\right)=\underset{{cyc}} {\sum}\frac{{a}}{{b}^{\mathrm{2}} +{c}} \\ $$$$\mathcal{L}\left({a},{b},{c},\lambda\right)={F}\left({a},{b},{c}\right)+\lambda\left({a}+{b}+{c}−\mathrm{3}\right) \\ $$$$\frac{\partial\mathcal{L}}{\partial{a}}=\frac{\mathrm{1}}{{b}^{\mathrm{2}} +{c}}−\frac{{b}}{\left({c}^{\mathrm{2}} +{a}^{\mathrm{2}} \right)^{\mathrm{2}} }−\frac{\mathrm{2}{ac}}{\left({a}^{\mathrm{2}} +{b}\right)^{\mathrm{2}} }+\lambda=\mathrm{0} \\ $$$$\frac{\partial\mathcal{L}}{\partial{b}}=\frac{\mathrm{1}}{{c}^{\mathrm{2}} +{a}}−\frac{{c}}{\left({a}^{\mathrm{2}} +{b}\right)^{\mathrm{2}} }−\frac{\mathrm{2}{ba}}{\left({b}^{\mathrm{2}} +{c}^{\mathrm{2}} \right)}+\lambda=\mathrm{0} \\ $$$$\frac{\partial\mathcal{L}}{\partial{c}}=\frac{\mathrm{1}}{{a}^{\mathrm{2}} +{b}}−\frac{{a}}{\left({b}^{\mathrm{2}} +{c}\right)^{\mathrm{2}} }−\frac{\mathrm{2}{ab}}{\left({c}^{\mathrm{2}} +{a}\right)^{\mathrm{2}} }+\lambda=\mathrm{0} \\ $$$${a}={b}={c},{a}+{b}+{c}=\mathrm{3}\Rightarrow{a}={b}={c}=\mathrm{1} \\ $$$$\lambda=−\frac{\mathrm{1}}{\mathrm{4}} \\ $$$${H}_{\mathcal{L}} \left(\mathrm{1},\mathrm{1},\mathrm{1}\right)=\begin{pmatrix}{\frac{\partial^{\mathrm{2}} {F}}{\partial{a}^{\mathrm{2}} }}&{\frac{\partial^{\mathrm{2}} {F}}{\partial{a}\partial{b}}}&{\frac{\partial^{\mathrm{2}} {F}}{\partial{a}\partial{c}}}\\{\frac{\partial^{\mathrm{2}} {F}}{\partial{b}\partial{a}}}&{\frac{\partial^{\mathrm{2}} {F}}{\partial{b}^{\mathrm{2}} }}&{\frac{\partial^{\mathrm{2}} {F}}{\partial{b}\partial{c}}}\\{\frac{\partial^{\mathrm{2}} {F}}{\partial{c}\partial{b}}}&{\frac{\partial^{\mathrm{2}} {F}}{\partial{c}\partial{b}}}&{\frac{\partial^{\mathrm{2}} {F}}{\partial{c}^{\mathrm{2}} }}\end{pmatrix}_{\left(\mathrm{1},\mathrm{1},\mathrm{1}\right)} \\ $$$$\frac{\partial^{\mathrm{2}} {F}}{\partial{a}^{\mathrm{2}} }\mid_{\left(\mathrm{1},\mathrm{1},\mathrm{1}\right)} =\frac{\mathrm{3}}{\mathrm{4}},\frac{\partial^{\mathrm{2}} {F}}{\partial{a}\partial{b}}\mid_{\left(\mathrm{1},\mathrm{1},\mathrm{1}\right)} =\frac{\mathrm{1}}{\mathrm{4}} \\ $$$${H}=\begin{pmatrix}{\frac{\mathrm{3}}{\mathrm{4}}}&{−\frac{\mathrm{1}}{\mathrm{4}}}&{−\frac{\mathrm{1}}{\mathrm{4}}}\\{−\frac{\mathrm{1}}{\mathrm{4}}}&{\frac{\mathrm{3}}{\mathrm{4}}}&{−\frac{\mathrm{1}}{\mathrm{4}}}\\{−\frac{\mathrm{1}}{\mathrm{4}}}&{−\frac{\mathrm{1}}{\mathrm{4}}}&{\frac{\mathrm{3}}{\mathrm{4}}}\end{pmatrix} \\ $$$$\mathrm{det}\begin{pmatrix}{\frac{\mathrm{3}}{\mathrm{4}}−\mu}&{−\frac{\mathrm{1}}{\mathrm{4}}}&{−\frac{\mathrm{1}}{\mathrm{4}}}\\{−\frac{\mathrm{1}}{\mathrm{4}}}&{\frac{\mathrm{3}}{\mathrm{4}}−\mu}&{−\frac{\mathrm{1}}{\mathrm{4}}}\\{−\frac{\mathrm{1}}{\mathrm{4}}}&{−\frac{\mathrm{1}}{\mathrm{4}}}&{\frac{\mathrm{3}}{\mathrm{4}}−\mu}\end{pmatrix}=\mathrm{0}\Rightarrow\left(\mu−\frac{\mathrm{1}}{\mathrm{4}}\right)\left(\mu−\mathrm{1}\right)^{\mathrm{2}} =\mathrm{0} \\ $$$$\sigma\left({H}\right)=\left\{\frac{\mathrm{1}}{\mathrm{4}}\mathrm{1},\mathrm{1}\right\}\subset\mathbb{R}^{+} \\ $$$$\therefore{H}\succ\mathrm{0}\left(\mathrm{1},\mathrm{1},\mathrm{1}\right)\mathrm{Is}\:\mathrm{a}\:\mathrm{strictly}\:\mathrm{local}\:\mathrm{minimum}\:\mathrm{point} \\ $$$$\underset{{S}} {\mathrm{min}}{F}={F}\left(\mathrm{1},\mathrm{1},\mathrm{1}\right)=\frac{\mathrm{1}}{\mathrm{1}^{\mathrm{2}} +\mathrm{1}}+\frac{\mathrm{1}}{\mathrm{1}^{\mathrm{2}} +\mathrm{1}}+\frac{\mathrm{1}}{\mathrm{1}^{\mathrm{2}} +\mathrm{1}}=\frac{\mathrm{3}}{\mathrm{2}} \\ $$$$\therefore\forall\left({a},{b},{c}\right)\in{S},\underset{{cyc}} {\sum}\frac{{a}}{{b}^{\mathrm{2}} −{c}}\geq\frac{\mathrm{3}}{\mathrm{2}} \\ $$

Commented by hardmath last updated on 01/Jan/26

$$\mathrm{Cooll}\:\mathrm{dear}\:\mathrm{professor}\:\mathrm{thank}\:\mathrm{you} \\ $$

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