How find this minimum $a+b+c+\sqrt{a^2+b^2}+\sqrt{b^2+c^2}+\sqrt{a^2+c^2}$

Question

let $a,b,c>0$ and such $$\dfrac{1}{a}+\dfrac{2}{b}+\dfrac{3}{c}=1$$.

find this follow minimum $$a+b+c+\sqrt{a^2+b^2}+\sqrt{b^2+c^2}+\sqrt{a^2+c^2}$$

I have know this How to find the minimum of $a+b+\sqrt{a^2+b^2}$

I want use Cauchy-Schwarz inequality,But I can't,

Thank you

See the Wikipedia article Lagrange multipler, and try to solve your problem that way. If you don't succeed, come back and tell us. — TonyK, Dec 05 '13 at 10:52
No,This inequality use Larange multipler is not well methods — math110, Dec 05 '13 at 11:13
Can you find the equilibrium point and apply the Cauchy method used in the previous question? — Calvin Lin, Dec 05 '13 at 15:09

score 0 · Accepted Answer · answered Dec 05 '13 at 16:53

We can follow a similar approach as to the question in your link, even though apriori we do not know when the minimum is obtained. Let the objective $P = a + b + c + \sqrt{a^2+b^2} + \sqrt{b^2+c^2} + \sqrt{a^2+c^2}$ attain its minimum when $a:b:c = x:y:1$

Now we use Cauchy-Schwarz, to establish $P \ge Q \ge R$ where $R$ is independent of $a, b, c$, while taking care equality is obtained when the above ratios are maintained. Down the line, we obtain conditions which allow us to determine $x, y$ as well.

By Cauchy-Schwarz: $$(a^2+b^2)(x^2+y^2) \ge (ax+by)^2 \implies \sqrt{a^2+b^2} \ge \dfrac{ax+by}{\sqrt{x^2+y^2}} $$

Similarly we have $\sqrt{b^2+c^2} \ge \dfrac{by+c}{\sqrt{y^2+1}}, \sqrt{c^2+a^2} \ge \dfrac{c+ax}{\sqrt{1+x^2}}$.

\begin{align} \therefore P &\ge (a+b+c) +\dfrac{ax+by}{\sqrt{x^2+y^2}} + \dfrac{by+c}{\sqrt{y^2+1}} + \dfrac{c+ax}{\sqrt{1+x^2}} \\ &= a\left(1+\frac{x}{\sqrt{x^2+y^2}} +\frac{x}{\sqrt{x^2+1}} \right)+ b\left(1+\frac{y}{\sqrt{x^2+y^2}} +\frac{y}{\sqrt{y^2+1}} \right) + c\left(1+\frac1{\sqrt{x^2+1}} +\frac1{\sqrt{y^2+1}} \right) \\ &=Q \end{align}

Now we can use C-S again and the constraint to establish a lower bound for $Q$ independent of $a, b, c$. As this bound is in fact attained during equality, this must necessarily be the minimum.

\begin{align} Q &= Q\left(\frac1a+\frac2b+\frac3c \right) \\ &\ge \left( \sqrt{1+\frac{x}{\sqrt{x^2+y^2}} +\frac{x}{\sqrt{x^2+1}}} + \sqrt{2\left(1+\frac{y}{\sqrt{x^2+y^2}} +\frac{y}{\sqrt{y^2+1}}\right)} + \sqrt{3\left(1+\frac1{\sqrt{x^2+1}} +\frac1{\sqrt{y^2+1}}\right) } \right)^2 \\ = R \end{align}

For the point of equality to be the same as that for $P \ge Q$, we must also have \begin{align} x^2 \left(1+\dfrac{x}{\sqrt{x^2+y^2}} +\dfrac{x}{\sqrt{x^2+1}} \right) &= \frac{y^2}2 \left(1+\dfrac{y}{\sqrt{x^2+y^2}} +\dfrac{y}{\sqrt{y^2+1}} \right) \\ &= \frac13 \left(1+\dfrac1{\sqrt{x^2+1}} +\dfrac1{\sqrt{y^2+1}} \right) \end{align}

So in theory we can solve for $x, y$ from equations above, to substitute in the expression for $R$, to find the value of the minimum. Well, in this particular case I am not so sure if finding exact values for $x, y$ is easy though...

Downvotes? Anyone care to explain? – Macavity Feb 27 '18 at 18:58 — Macavity, Feb 27 '18 at 18:58

How find this minimum $a+b+c+\sqrt{a^2+b^2}+\sqrt{b^2+c^2}+\sqrt{a^2+c^2}$

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