Find minimum of $a+b+c+\frac1a+\frac1b+\frac1c$ given that: $a+b+c\le \frac32$

Question

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?!

Answer 1

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. $a=b=c=\frac{x}{3}=\frac{1}{2}\,$.

Answer 2

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!

Answer 3

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 $t=3$

Answer 4

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.

Answer 5

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 $a=b=c=S/3=1/2.$