4
$\begingroup$

Find minimum of $a+b+c+\frac1a+\frac1b+\frac1c$ given that: $a+b+c\le \frac32$ ($a,b,c$ are positive real numbers).

There is a solution, which relies on guessing the minimum case happening at $a=b=c=\frac12$ and then applying AM-GM inequality,but what if one CANNOT guess that?!

$\endgroup$
  • $\begingroup$ $f(x)=x+\frac1x$ is a convex function when $x \gt 0$ $\endgroup$ – Henry Jul 16 '17 at 19:24
5
$\begingroup$

Hint:  by the AM-HM (arithmetic-harmonic mean) inequality:

$$ \frac{a+b+c}{3} \ge \frac{3}{\frac{1}{a}+\frac{1}{b}+\frac{1}{c}} \;\;\iff\;\; \frac{1}{a}+\frac{1}{b}+\frac{1}{c} \ge \frac{9}{a+b+c} $$

Let $x=a+b+c \in (0,\frac{3}{2}]\,$, then the expression to be minimized can be written as:

$$ a+b+c+\frac1a+\frac1b+\frac1c \ge a+b+c+\frac{9}{a+b+c} = x + \frac{9}{x} $$

The function $f(x)=x + \frac{9}{x}$ is decreasing on $(0,\frac{3}{2}]\,$, so $f(x) \ge f(\frac{3}{2})=\frac{15}{2}\,$ for $x \in (0,\frac{3}{2}]$.

The minimum value of $\frac{15}{2}$ is attained when $x=\frac{3}{2}$ and AM=HM i.e. $a=b=c=\frac{x}{3}=\frac{1}{2}\,$.

$\endgroup$
  • $\begingroup$ Suppose that in this question there is no clue to the minimum value(i.e. 15/2).Can you offer a Brand New solution from scratch? $\endgroup$ – Hamid Reza Ebrahimi Jul 16 '17 at 20:19
  • $\begingroup$ @HamidRezaEbrahimi This answer does not pre-suppose any particular solution. The lower bound of $15/2$ is determined from the respective inequalities, the fact that it is attained for $x=3/2$ is a direct consequence of the monotonicity of $f(x)\,$, and $\,a=b=c\,$ follows from the equality case of AM-HM. So, again, the above does not rely on any guesswork or prior knowledge of any of those values, it derives them from scratch. $\endgroup$ – dxiv Jul 16 '17 at 20:24
3
$\begingroup$

For $a=b=c=\frac{1}{2}$ we get a value $\frac{15}{2}$.

We'll prove that it's a minimal value.

Indeed, we need to prove that $$\sum_{cyc}\left(a+\frac{1}{a}-\frac{5}{2}\right)\geq0$$ or $$\sum_{cyc}\frac{(a-2)(2a-1)}{a}\geq0$$ or $$\sum_{cyc}\left(\frac{(2a-1)(a-2)}{a}+3(2a-1)\right)+6\left(\frac{3}{2}-a-b-c\right)\geq0$$ or $$\sum_{cyc}\frac{2(2a-1)^2}{a}+6\left(\frac{3}{2}-a-b-c\right)\geq0,$$ which is obvious.

Done!

$\endgroup$
  • $\begingroup$ Please note that the question wants a different way for finding the minimum,NOT proving the minimum at $a=b=c=\frac12$ $\endgroup$ – Hamid Reza Ebrahimi Jul 16 '17 at 19:27
  • $\begingroup$ @Hamid Reza Ebrahimi I fixed again. $\endgroup$ – Michael Rozenberg Jul 16 '17 at 20:06
  • $\begingroup$ Suppose that in this question there is no clue to the minimum value(i.e. 15/2).Can you offer a Brand New solution from scratch? $\endgroup$ – Hamid Reza Ebrahimi Jul 16 '17 at 20:13
1
$\begingroup$

your inequality is equivalent to $$2\left(\frac{1}{a}+\frac{1}{b}+\frac{1}{c}\right)+2(a+b+c)\geq 15$$ By $AM-HM$ we get $$a+b+c\geq \frac{9}{\frac{1}{a}+\frac{1}{b}+\frac{1}{c}}$$ thus $$2(a+b+c)\geq \frac{18}{\frac{1}{a}+\frac{1}{b}+\frac{1}{c}}$$ thus we have $$2\left(\frac{1}{a}+\frac{1}{b}+\frac{1}{c}\right)+\frac{18}{\frac{1}{a}+\frac{1}{b}+\frac{1}{c}}\geq 15$$ Setting $$t=\frac{1}{a}+\frac{1}{b}+\frac{1}{c}$$ so we have to prove $$2t+\frac{18}{t}\geq 15$$ this is equivalent to $$2t^2-15t+18\geq 0$$ or $$t\le \frac{3}{2}$$ or $$t\geq 6$$ or we can consider the function $$f(t)=2t+\frac{18}{t}$$ and $$f'(t)=2-\frac{18}{t^2}$$ and $$f'(t)=0$$ for $t=3$

$\endgroup$
1
$\begingroup$

WLOG $a\geq b\geq c.$

Let $f(a,b,c)=a+b+c+1/a+/b+1/c.$

For a given value of $a+b+c,$ let $c$ remain constant while $a,b$ vary , subject to the constraint that $a+b$ is constant, so that $a+b+c$ also remains constant. Then $db/da=-1$ and $d(1/b)/da=1/a^2 .$ So with constant $c$ we have $$df(a,b,c)/da=-1/a^2+1/b^2=(a-b)(a+b)/a^2b^2\geq 0$$ (because $a\geq b$). So we cannot have a minimum of $f(a,b,c)$ for a given value of $a+b+c$ unless $a=b.$

Applying this method again, leaving $a$ constant and letting $b,c$ vary, subject to the constraint that $b+c$ is constant, we see also that we cannot have a minimum of $f$ for a given value of $a+b+c$ unless $b=c.$

Therefore for each $S\in (0,3/2]$ we have $$\min \{f(a,b,c): a+b+c=S\}=f(S/3,S/3,S/3)=S+9/S.$$ The least value of $S+9/S$ for $S\in (0,3/2]$, is $15/2,$ which occurs uniquely at $S= 3/2.$ And as we have seen , this only occurs when $a=b=c=S/3=1/2.$

$\endgroup$
0
$\begingroup$

You may use Lagrange multipliers method to show that we need $a=b=c$. The optimisation problem would be to minimise $a+b+c+\frac{1}{a}+\frac{1}{b}+\frac{1}{c}$, having the constraint $a+b+c=t , t\leq \frac{3}{2}$.

$$F=a+b+c+\frac{1}{a}+\frac{1}{b}+\frac{1}{c}+\lambda(a+b+c-t)$$

Then, we need to find the right $t$, for which the minimum is achieved. Knowing that $a=b=c$, the problem is simplified to minimizing

$$3(a+\frac{1}{a})$$

As mentioned in comments, it is a convex problem and it achieves its minimum at $a=1$. However, the constraint $a+b+c\leq\frac{3}{2}$ does not allow $a=1$. So, because of convexity, you should decrease $a$ until you satisfy the constraint.

$\endgroup$

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.