3

Prove that for all positive real numbers: $$\dfrac{a}{b+c}+\dfrac{b}{c+a}+\dfrac{c}{a+b}\geq\dfrac{3}{2}$$

This is same as this question but a different approach is used there whereas I want to verify my approach to this problem.

My Approach:

$$\dfrac{a}{b+c}+\dfrac{b}{c+a}+\dfrac{c}{a+b}=\Big(\dfrac{a}{b+c}+1\Big)+\Big(\dfrac{b}{c+a}+1\Big)+\Big(\dfrac{c}{a+b}+1\Big)-3$$ $$=(a+b+c)\Big[\dfrac{1}{b+c}+\dfrac{1}{c+a}+\dfrac{1}{a+b}\Big]-3$$ By AM-HM inequality: $$\dfrac{3}{\dfrac{1}{b+c}+\dfrac{1}{c+a}+\dfrac{1}{a+b}}\leq\dfrac{2(a+b+c)}{3}\Rightarrow (a+b+c)\Big[\dfrac{1}{b+c}+\dfrac{1}{c+a}+\dfrac{1}{a+b}\Big]\geq \dfrac{9}{2}$$ $$(a+b+c)\Big[\dfrac{1}{b+c}+\dfrac{1}{c+a}+\dfrac{1}{a+b}\Big]-3\geq \dfrac{3}{2}$$ $\therefore \dfrac{a}{b+c}+\dfrac{b}{c+a}+\dfrac{c}{a+b}\geq\dfrac{3}{2}\space \forall\ a,b,c\in \mathbb R$ and $a,b,c>0$

Please check this approach and provide suggestions. Also please provide alternative solutions if available.

THANKS

Michael Rozenberg
  • 194,933
Devansh Kamra
  • 1,808
  • 2
    your proof is fine this called nesbitt inequality.for more proofs https://en.wikipedia.org/wiki/Nesbitt%27s_inequality – Albus Dumbledore Sep 01 '20 at 11:02
  • Wouldn't it be good to post it as answer to that question rather than separated post, and see that if it will get upvoted or commented or not? – C.F.G Sep 01 '20 at 11:22
  • @C.F.G, not sure if the question has been active for a long time now. Plus the question I attached was not about the inequality itself but to get to a step in a specific approach whereas I want to check whether my approach is correct or not. – Devansh Kamra Sep 01 '20 at 11:24

4 Answers4

2

Your solution is right.

Also, SOS helps: $$\sum_{cyc}\frac{a}{b+c}-\frac{3}{2}=\sum_{cyc}\left(\frac{a}{b+c}-\frac{1}{2}\right)=\sum_{cyc}\frac{2a-b-c}{2(b+c)}=$$ $$=\sum_{cyc}\frac{a-b-(c-a)}{2(b+c)}=\sum_{cyc}\left(\frac{a-b}{2(b+c)}-\frac{c-a}{2(b+c)}\right)=$$ $$=\sum_{cyc}\left(\frac{a-b}{2(b+c)}-\frac{a-b}{2(c+a)}\right)=\sum_{cyc}(a-b)\left(\frac{1}{2(b+c)}-\frac{1}{2(c+a)}\right)=$$ $$=\sum_{cyc}\frac{(a-b)^2}{2(a+c)(b+c)}\geq0.$$ Now we see that the starting inequality is true for any reals $a$, $b$ and $c$ such that $ab+ac+bc>0.$

Also, there is a solution by AM-GM, by C-S, by TL, by $uvw$ and by more and more and more.

Michael Rozenberg
  • 194,933
1

Solution by Buffalo Way method.

Let $a=\min\{a,b,c\},$ $b=a+u$ and $c=a+v$.

Thus, $$2\prod_{cyc}(a+b)\left(\sum_{cyc}\frac{a}{b+c}-\frac{3}{2}\right)=4(u^2-uv+v^2)a+(u+v)(2u^2-3uv+2v^2)\geq0.$$

Michael Rozenberg
  • 194,933
0

Solution by the Tangent Line method.

Since our inequality is homogeneous, we can assume that $a+b+c=3$ and we onbtain: $$\sum_{cyc}\frac{a}{b+c}-\frac{3}{2}=\sum_{cyc}\frac{a}{3-a}-\frac{3}{2}=\sum_{cyc}\left(\frac{a}{3-a}-\frac{1}{2}\right)=$$ $$\sum_{cyc}\frac{3(a-1)}{2(3-a)}=\frac{3}{2}\sum_{cyc}\left(\frac{a-1}{3-a}-\frac{1}{2}(a-1)\right)=\frac{9}{4}\sum_{cyc}\frac{(a-1)^2}{3-a}\geq0.$$

Michael Rozenberg
  • 194,933
0

Another proof$:$

Due to homogeneous, assume $a+b+c=1.$

Let $p=a+b+c=1,q=\dfrac{1-t^2}{3} \quad(\, t\in [\,0,1\,]\,),r=abc.$

Need to prove$:$ $$\frac73\,{t}^{2}+9\,r-\frac13 \geqslant 0$$

Since $$r\geqslant \dfrac{1}{27} \left( 1-2t \right) \left( 1+t \right) ^{2}$$

We need to prove$:$ $$\dfrac{2}{3} t^2(2-t) \geqslant 0,$$

which is true since $t \in [\,0,\,1\,].$

See also here.

There is also a proof by SS (SOS - Schur) method.

$$\text{LHS}-\text{RHS}={\frac {2\, \left( a-b \right) ^{2} \left( a+b \right) + \left( a -c \right) \left( b-c \right) \left( a+b+2\,c \right) }{2 \left( b+c \right) \left( c+a \right) \left( a+b \right) }} \geqslant 0,$$ which is obvious if $c\equiv \min\{a,b,c\}.$

NKellira
  • 2,001