Question Number 76061 by tw000001 last updated on 23/Dec/19
$$\mathrm{What}'\mathrm{s}\:\mathrm{the}\:\mathrm{minimum}\:\mathrm{value}\:\mathrm{of} \\ $$$$\frac{\mathrm{13}{a}+\mathrm{13}{b}+\mathrm{2}{c}}{\mathrm{2}{a}+\mathrm{2}{b}}+\frac{\mathrm{24}{a}−{b}+\mathrm{13}{c}}{\mathrm{2}{b}+\mathrm{2}{c}}+\frac{−{a}+\mathrm{24}{b}+\mathrm{13}{c}}{\mathrm{2}{a}+\mathrm{2}{c}}? \\ $$$$\left({a},{b},{c}\:\mathrm{are}\:\mathrm{positive}\:\mathrm{numbers}.\right) \\ $$$$\mathrm{I}\:\mathrm{think}\:\mathrm{nobody}\:\mathrm{can}\:\mathrm{solve}\:\mathrm{this}. \\ $$
Answered by MJS last updated on 23/Dec/19
![((13a+13b+2c)/(2a+2b))+((24a−b+13c)/(2b+2c))+((−a+24b+13c)/(2a+2c))= =(c/(a+b))+((12b+7c)/(a+c))+((12a+7c)/(b+c))+((11)/2) we see that a and b are of equal weight but c is different let a=pc∧b=qc (1/(p+q))+((12q+7)/(p+1))+((12p+7)/(q+1))+((11)/2) to ensure positive values let p=x^2 ∧q=y^2 (1/(x^2 +y^2 ))+((12y^2 +7)/(x^2 +1))+((12x^2 +7)/(y^2 +1))+((11)/2) now I simply tried y=x −((10)/(x^2 +1))+(1/(2x^2 ))+((59)/2)=((59x^4 +40x^2 +1)/(2x^2 (x^2 +1))) (d/dx)[((59x^4 +40x^2 +1)/(2x^2 (x^2 +1)))]=0 ((19x^4 −2x^2 −1)/(x^3 (x^2 +1)^2 ))=0 ⇒ x^2 =((1+2(√5))/(19)) ⇒ min ((59x^4 +40x^2 +1)/(2x^2 (x^2 +1))) =19+2(√5)≈23.472136 at a=b=((1+2(√5))/(19))c; c∈R^+](https://www.tinkutara.com/question/Q76066.png)
$$\frac{\mathrm{13}{a}+\mathrm{13}{b}+\mathrm{2}{c}}{\mathrm{2}{a}+\mathrm{2}{b}}+\frac{\mathrm{24}{a}−{b}+\mathrm{13}{c}}{\mathrm{2}{b}+\mathrm{2}{c}}+\frac{−{a}+\mathrm{24}{b}+\mathrm{13}{c}}{\mathrm{2}{a}+\mathrm{2}{c}}= \\ $$$$=\frac{{c}}{{a}+{b}}+\frac{\mathrm{12}{b}+\mathrm{7}{c}}{{a}+{c}}+\frac{\mathrm{12}{a}+\mathrm{7}{c}}{{b}+{c}}+\frac{\mathrm{11}}{\mathrm{2}} \\ $$$$\mathrm{we}\:\mathrm{see}\:\mathrm{that}\:{a}\:\mathrm{and}\:{b}\:\mathrm{are}\:\mathrm{of}\:\mathrm{equal}\:\mathrm{weight}\:\mathrm{but} \\ $$$${c}\:\mathrm{is}\:\mathrm{different} \\ $$$$\mathrm{let}\:{a}={pc}\wedge{b}={qc} \\ $$$$\frac{\mathrm{1}}{{p}+{q}}+\frac{\mathrm{12}{q}+\mathrm{7}}{{p}+\mathrm{1}}+\frac{\mathrm{12}{p}+\mathrm{7}}{{q}+\mathrm{1}}+\frac{\mathrm{11}}{\mathrm{2}} \\ $$$$\mathrm{to}\:\mathrm{ensure}\:\mathrm{positive}\:\mathrm{values} \\ $$$$\mathrm{let}\:{p}={x}^{\mathrm{2}} \wedge{q}={y}^{\mathrm{2}} \\ $$$$\frac{\mathrm{1}}{{x}^{\mathrm{2}} +{y}^{\mathrm{2}} }+\frac{\mathrm{12}{y}^{\mathrm{2}} +\mathrm{7}}{{x}^{\mathrm{2}} +\mathrm{1}}+\frac{\mathrm{12}{x}^{\mathrm{2}} +\mathrm{7}}{{y}^{\mathrm{2}} +\mathrm{1}}+\frac{\mathrm{11}}{\mathrm{2}} \\ $$$$\mathrm{now}\:\mathrm{I}\:\mathrm{simply}\:\mathrm{tried}\:{y}={x} \\ $$$$−\frac{\mathrm{10}}{{x}^{\mathrm{2}} +\mathrm{1}}+\frac{\mathrm{1}}{\mathrm{2}{x}^{\mathrm{2}} }+\frac{\mathrm{59}}{\mathrm{2}}=\frac{\mathrm{59}{x}^{\mathrm{4}} +\mathrm{40}{x}^{\mathrm{2}} +\mathrm{1}}{\mathrm{2}{x}^{\mathrm{2}} \left({x}^{\mathrm{2}} +\mathrm{1}\right)} \\ $$$$\frac{{d}}{{dx}}\left[\frac{\mathrm{59}{x}^{\mathrm{4}} +\mathrm{40}{x}^{\mathrm{2}} +\mathrm{1}}{\mathrm{2}{x}^{\mathrm{2}} \left({x}^{\mathrm{2}} +\mathrm{1}\right)}\right]=\mathrm{0} \\ $$$$\frac{\mathrm{19}{x}^{\mathrm{4}} −\mathrm{2}{x}^{\mathrm{2}} −\mathrm{1}}{{x}^{\mathrm{3}} \left({x}^{\mathrm{2}} +\mathrm{1}\right)^{\mathrm{2}} }=\mathrm{0}\:\Rightarrow\:{x}^{\mathrm{2}} =\frac{\mathrm{1}+\mathrm{2}\sqrt{\mathrm{5}}}{\mathrm{19}} \\ $$$$\Rightarrow\:\mathrm{min}\:\frac{\mathrm{59}{x}^{\mathrm{4}} +\mathrm{40}{x}^{\mathrm{2}} +\mathrm{1}}{\mathrm{2}{x}^{\mathrm{2}} \left({x}^{\mathrm{2}} +\mathrm{1}\right)}\:=\mathrm{19}+\mathrm{2}\sqrt{\mathrm{5}}\approx\mathrm{23}.\mathrm{472136} \\ $$$$\mathrm{at}\:{a}={b}=\frac{\mathrm{1}+\mathrm{2}\sqrt{\mathrm{5}}}{\mathrm{19}}{c};\:{c}\in\mathbb{R}^{+} \\ $$
Commented by MJS last updated on 23/Dec/19
btw my name is Nobody ��
Commented by peter frank last updated on 23/Dec/19
$${hahahahah}\:{great}\:{man}\:{mr}\:{mjs} \\ $$