Nice_AM-GM

[dduclam]

Let [tex]a,b,c>0[/tex] such that [tex]abc=1[/tex]. Prove that

[tex]\frac1{a^2+3bc}+\frac1{b^2+3ca}+\frac1{c^2+3ab}\le \frac3{4}[/tex]


dduclam

[redmafiya]

[tex]L \eq \frac{1}{a^2+3bc } + \frac{1}{b^2+3ca} + \frac{1}{c^2+3ab }[/tex]
[tex]\eq \frac{a}{a^3+3abc } + \frac{b}{b^3+3bca} + \frac{c}{c^3+3cab }[/tex]
[tex]\eq \frac{a}{a^3+3} + \frac{b}{b^3+3} + \frac{c}{c^3+3}[/tex]

As [tex]a^3+3 \eq (a^3 + 1 + 1) + 1[/tex]
Applying AM - GM Theorem, we have
[tex]a^3+3 \eq (a^3 + 1 + 1) + 1 \geq 3a + 1[/tex]

For the same reason:
[tex]b^3+3 \eq (b^3 + 1 + 1) + 1 \geq 3b + 1[/tex]
[tex]c^3+3 \eq (c^3 + 1 + 1) + 1 \geq 3c + 1[/tex]

So
[tex]L \leq \frac{a}{3a+1 } + \frac{b}{3b+1 } + \frac{c}{ 3c+1}[/tex]

Next we need to show that
[tex]\frac{a}{3a+1 } + \frac{b}{3b+1 } + \frac{c}{ 3c+1} \leq \frac{3}{4}[/tex]

OR to show that

[tex]4[a(3b+1)(3c+1)+b(3a+1)(3c+1)+c(3a+1)(3b+1)] -3(3a+1)(3b+1)(3c+1) \leq 0[/tex]

In fact, simplifying and applying AM-GM theorem again, we have

[tex]4[a(3b+1)(3c+1)+b(3a+1)(3c+1)+c(3a+1)(3b+1)] -3(3a+1)(3b+1)(3c+1)[/tex]
[tex]\eq 24-3(ab+bc+ac)-5(a+b+c)[/tex]
[tex]\leq 24-3(3)-5(3)[/tex]
[tex]\eq 0[/tex]

[Math Tutor]

Excellent solution redmafiya !

[dduclam]

[tex]L \eq \frac{1}{a^2+3bc } + \frac{1}{b^2+3ca} + \frac{1}{c^2+3ab }[/tex]
[tex]\eq \frac{a}{a^3+3abc } + \frac{b}{b^3+3bca} + \frac{c}{c^3+3cab }[/tex]
[tex]\eq \frac{a}{a^3+3} + \frac{b}{b^3+3} + \frac{c}{c^3+3}[/tex]

As [tex]a^3+3 \eq (a^3 + 1 + 1) + 1[/tex]
Applying AM - GM Theorem, we have
[tex]a^3+3 \eq (a^3 + 1 + 1) + 1 \geq 3a + 1[/tex]

For the same reason:
[tex]b^3+3 \eq (b^3 + 1 + 1) + 1 \geq 3b + 1[/tex]
[tex]c^3+3 \eq (c^3 + 1 + 1) + 1 \geq 3c + 1[/tex]

So
[tex]L \leq \frac{a}{3a+1 } + \frac{b}{3b+1 } + \frac{c}{ 3c+1}[/tex]

Next we need to show that
[tex]\frac{a}{3a+1 } + \frac{b}{3b+1 } + \frac{c}{ 3c+1} \leq \frac{3}{4}[/tex]

OR to show that

[tex]4[a(3b+1)(3c+1)+b(3a+1)(3c+1)+c(3a+1)(3b+1)] -3(3a+1)(3b+1)(3c+1) \leq 0[/tex]

In fact, simplifying and applying AM-GM theorem again, we have

[tex]4[a(3b+1)(3c+1)+b(3a+1)(3c+1)+c(3a+1)(3b+1)] -3(3a+1)(3b+1)(3c+1)[/tex]
[tex]\eq 24-3(ab+bc+ac)-5(a+b+c)[/tex]
[tex]\leq 24-3(3)-5(3)[/tex]
[tex]\eq 0[/tex]

Yeah,It's also my fist proof,by AM-GM. However,I still have some nother solutions,exam

By AM-GM we have [tex]a+b+c\ge3[/tex],we only consider the case [tex]a+b+c=3[/tex]
Then [tex]\frac a{a^3+3}\le \frac1{4}+\frac1{16}(a-1)[/tex] because it's equivalent to [tex]\frac1{4}(a-1)^2(a^2+5a+9)\ge0[/tex] always true.

Hence [tex]\frac{a}{a^3+3} + \frac{b}{b^3+3} + \frac{c}{c^3+3}\le \frac3{4}+\frac1{16}(a+b+c-3)=\frac3{4}[/tex]

We have done !

[dduclam]

With above inequality,we can use AM-GM. But with following harder inequality,we can't use AM-GM:

[tex]\frac{a}{a^2+3}+\frac{b}{b^2+3}+\frac{c}{c^2+3}\le\frac3{4};\forall a,b,c>0:abc=1[/tex]

Do you think same?

[MM]

Expansion:
[tex]\frac{3}{4}-\frac{a}{a^{2}+3}-\frac{b}{b^{2}+3}-\frac{c}{c^{2}+3}=\frac{9(a^{2}b^{2}+a^{2}c^{2}+b^{2}c^{2})+3(abc)^{2}-4abc(ab+ac+bc)-12(a^{2}b+a^{2}c+b^{2}a+b^{2}c+c^{2}a+c^{2}b)-36(a+b+c)+27(a^{2}+b^{2}+c^{2})+81}{4(a^{2}+3)(b^{2}+3)(c^{2}+3)}\ge 0[/tex]. It's enough to prove that the numerator is bigger than or equal to zero. Let [tex]a+b+c=p[/tex] and [tex]ab+ac+bc=q[/tex]. Using [tex]abc=1[/tex] we obtain that the numerator is equal to [tex]9(q^{2}-2p)+3-4q-12(pq-3)-36p+27(p^{2}-2q)+81[/tex]. Thus we obtain by AM-GM [tex]9q^{2}-54p+120-12pq+27q^{2}-58q\ge 9*3^{2}-54*3+120-12*3*3+27*3^{2}-58*3=0[/tex]. Is this true?