# Find min of $\frac{a}{b+c}+\frac{b}{a+c}+\frac{c}{a+b}+4\sqrt{2}\sqrt{\frac{ab+bc+ac}{a^2+b^2+c^2}}$ [closed]

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

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

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• 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? – Mohammad Zuhair Khan Jan 28 at 6:22
• Oh yes but I can't prove – Trương Văn Hào Jan 28 at 6:22
• 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 $6?$ – Mohammad Zuhair Khan Jan 28 at 6:26
• Can you prove it $\geq$ 6 ?? I think you correct . – Trương Văn Hào Jan 28 at 6:28
• 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. – Mohammad Zuhair Khan Jan 28 at 6:37

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: $$a^2+1+\sqrt{2a(a^2+1)}\geq2a+\sqrt{2a\cdot2a}=4a.$$

• Shouldn't it be true for all $a=b \gt 0$ and $c=0?$ – Mohammad Zuhair Khan Jan 28 at 7:02
• @Mohammad Zuhair Khan Yes, of course. I proved that it's enough to prove and I proved our inequality in this case. – Michael Rozenberg Jan 28 at 7:04

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.

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

• 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 $k=1$. – TheSimpliFire Mar 15 at 7:17