Suppose #0< a,b,c < 1# and #ab + bc + ca = 1#. Find the minimum value of #a + b + c + abc#?

2 Answers
May 13, 2017

#a+b+c+abc≥2#

Explanation:

By the arithmetic mean-geometric mean-harmonic mean inequality, #(a+b+c+abc)/4≥root(4)(a^2b^2c^2)≥4/(1/a+1/b+1/c+1/(abc))#, or #a+b+c+abc≥4sqrt(abc)≥16/(1/a+1/b+1/c+1/(abc))#.

Since #16/(1/a+1/b+1/c+1/(abc))=16/((bc+ac+ab+1)/(abc))# and #ab+bc+ac=1#, #a+b+c+abc≥4sqrt(abc)≥8abc#.

From #4sqrt(abc)≥8abc#, we can conclude that #abc≤1/4#. Then, #a+b+c+abc≥2#.

May 13, 2017

See below.

Explanation:

One way to elude the inequality restrictions is with a change of variable so making

#{(a->(sinalpha+1)/2),(b->(sinbeta+1)/2),(c->(singamma+1)/2):}#

the problem now in the variables #alpha, beta, gamma# has only equality restrictions and then we can approach it with the usual lagrange multipliers technique.

#f(alpha,beta,gamma) = 1/8(sinalpha+1)(sinbeta+1)(singamma+1)#
#g(alpha,beta,gamma)=1/4((sinalpha+1)(sinbeta+1)+(sinalpha+1)(singamma+1)+(sinbeta+1)(singamma+1))-1#

with the lagrangian

#L(alpha,beta,gamma,lambda)=f(alpha,beta,gamma)+lambda g(alpha,beta,gamma)#

The stationary point conditions are obtained by solving

#grad L=0#

or

#{(1/8 Cos alpha(5 + 4 lambda + Sin gamma + 2 lambda Sin gamma + Sin beta (1 + 2 lambda + Sin gamma))=0),(1/8 Cosbeta (5 + 4 lambda + Singamma + 2 lambda Singamma + Sinalpha(1 + 2 lambda + Singamma))=0),(1/8 Cosgamma(5 + 4 lambda + Sinbeta + 2 lambda Sinbeta + Sinalpha(1 + 2 lambda + Sinbeta))=0),(1/4 (2 Singamma + Sinbeta (2 + Singamma) + Sinalpha (2 + Sinbeta + Singamma)-1)=0):}#

Solving this system of equations with the help of Wolfram Alpha we get at

#((a,b,c,f(a,b,c)),(0, 1, 1,2),(1,0,1,2),(1,1,0,2),(1/sqrt[3], 1/sqrt[3], 1/sqrt[3],10/(3 sqrt[3])))#

Here #f(a,b,c) = a + b + c + a b c#

All this trouble could be avoided by observing the problem symmetry, so guessing #a=b=c#