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Given the three real numbers a, b, c are not negative, in which at most some are equal to zero. Find min of $\frac{a}{b+c}+\frac{b}{a+c}+\frac{c}{a+b}+4\sqrt{2}\sqrt{\frac{ab+bc+ac}{a^2+b^2+c^2}}$ Thanks so much

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closed as off-topic by RRL, José Carlos Santos, mrtaurho, Ali Caglayan, Trevor Gunn Jan 28 at 16:28

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  • $\begingroup$ Using $a=b=c$ gets me to $\frac 12+\frac 12+\frac 12+4\sqrt 2 \cdot \sqrt 1=4\sqrt 2+\frac 32.$ Can you check to see if there is a value smaller than this? $\endgroup$ – Mohammad Zuhair Khan Jan 28 at 6:22
  • $\begingroup$ Oh yes but I can't prove $\endgroup$ – Trương Văn Hào Jan 28 at 6:22
  • $\begingroup$ Setting $a=b$ and $c=0$ gives me $1+1+0+4\sqrt 2 \cdot \sqrt {\frac 12}=2+4=6 \lt 4\sqrt 2+\frac 32$ so that is clearly not the answer. Can you try other combinations to find something less than $6?$ $\endgroup$ – Mohammad Zuhair Khan Jan 28 at 6:26
  • $\begingroup$ Can you prove it $\geq$ 6 ?? I think you correct . $\endgroup$ – Trương Văn Hào Jan 28 at 6:28
  • $\begingroup$ I think you meant that the $\text {min} \leq 6$. Yes, because I have already found a value that is $6$, the minimum could either be that or something larger. I will try a more mathematical proof. $\endgroup$ – Mohammad Zuhair Khan Jan 28 at 6:37
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For $a=b=1$ and $c=0$ we get a value $6$.

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

Indeed, let $a+b+c=3u$, $ab+ac+bc=3v^2$ and $abc=w^3$.

Thus, we need to prove that $$\sum_{cyc}\frac{a}{b+c}+4\sqrt{\frac{2v^2}{3u^2-2v^2}}\geq6$$ or $$\frac{\sum\limits_{cyc}(a^3+a^2b+a^2c+abc)}{\prod\limits_{cyc}(a+b)}+4\sqrt{\frac{2v^2}{3u^2-2v^2}}\geq6$$ or $$\frac{3u(9u^2-6v^2)+3w^3}{9uv^2-w^3}+4\sqrt{\frac{2v^2}{3u^2-2v^2}}\geq6,$$ which is $f(w^3)\geq0,$ where $f$ is a linear function.

But the linear function gets a minimal value for the extreme value of $w^3$,

which happens in the following cases.

  1. Two variables are equal.

Let $b=c=1$.

Thus, we need to prove that $$\frac{a}{2}+\frac{2}{a+1}+4\sqrt{\frac{2(2a+1)}{a^2+2}}\geq6,$$ which is smooth;

  1. $b=1$ and $c=0$.

We need to prove that $$a+\frac{1}{a}+4\sqrt2\sqrt{\frac{a}{a^2+1}}\geq6,$$ which is smooth too:

$$a+\frac{1}{a}-2\geq4\left(1-\sqrt{\frac{2a}{a^2+1}}\right)$$ or $$\frac{1}{a}\geq\frac{4}{\sqrt{a^2+1}\left(\sqrt{a^2+1}+\sqrt{2a}\right)}$$ or $$a^2+1+\sqrt{2a(a^2+1)}\geq4a,$$ which is true by AM-GM: $$a^2+1+\sqrt{2a(a^2+1)}\geq2a+\sqrt{2a\cdot2a}=4a.$$

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  • $\begingroup$ Shouldn't it be true for all $a=b \gt 0$ and $c=0?$ $\endgroup$ – Mohammad Zuhair Khan Jan 28 at 7:02
  • $\begingroup$ @Mohammad Zuhair Khan Yes, of course. I proved that it's enough to prove and I proved our inequality in this case. $\endgroup$ – Michael Rozenberg Jan 28 at 7:04
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Setting $c=0$, we find that it has simplified to finding $\text{min } \frac ab+ \frac ba +4\sqrt 2 \sqrt {\frac{ab}{a^2+b^2}}$

Assuming $a \geq b$, we set $a=kb$ to obtain $k+\frac 1k+4\sqrt 2 \sqrt {\frac {k}{k^2+1}}=\frac {k^2+1}k+4\sqrt 2 \sqrt {\frac k{k^2+1}}$

As we have to find $\text {min} \frac {k^2+1}k+4\sqrt 2 \sqrt {\frac k{k^2+1}},$ we could try and figure out the minimum of $f(k)=\frac {k^2+1}k+4\sqrt 2 \sqrt {\frac k{k^2+1}}$ by finding $k$ when $f'(k)=0.$ However, being not skilled enough in that aspect, I opted to graph it and find the minimum point.

graph

A we can see, the minimum of $f(k)=6$ occurs when $k=1,$ giving us a minimum when $a=b$ and $c=0.$

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  • 1
    $\begingroup$ Note that $$f'(k)=1-\frac1{k^2}-\frac{2\sqrt2}{\sqrt{\frac{k}{k^2+1}}}\cdot\frac{1(k^2+1)-k(2k)}{(k^2+1)^2}=0\implies\frac{2\sqrt2(k^2-1)}{(k^2+1)^2}\sqrt{\frac{k^2+1}k}=1-\frac1{k^2}$$ giving $$\frac{k^2-1}{(k^2+1)^2}\sqrt{\frac{k^2+1}k}=\frac1{2\sqrt2}\left(1-\frac1{k^2}\right)\implies(k^2+1)^3=8k^3$$ after squaring both sides. Thus we have $$(k^2+1)^3-8k^3=0\implies(k-1)^2(k^4+2k^3+6k^2+2k+1)=0$$ and the quartic has no real solutions due to the positive coefficients and $k>0$. Therefore the only solution is $k=1$. $\endgroup$ – TheSimpliFire Mar 15 at 7:17

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