Inequalities with 3 and 4 variables

[MM]

If a,b,c>0 and a+b+c=3 prove that [tex]\frac{a}{b^{2}+1}+\frac{b}{c^{2}+1}+\frac{c}{a^{2}+1}\ge \frac{3}{2}[/tex]. When does the equality occur?
If a,b,c,d>0 and a+b+c+d=4 prove that [tex]\frac{a}{b^{2}+1}+\frac{b}{c^{2}+1}+\frac{c}{d^{2}+1}+\frac{d}{a^{2}+1}\ge 2[/tex]. When does the equality occur?
Is this inequality true for 5 or more varibales?

[redmafiya]

The proof for the first inequality is as follows:

[tex]\frac{a}{b^{2}+1}[/tex]

[tex]=\frac{a(b^{2}+1)-ab^{2}}{b^{2}+1}[/tex]

[tex]=a-\frac{ab^{2}}{b^{2}+1}[/tex]

[tex]\ge a-\frac{1}{2}ab[/tex]

Similarly
[tex]\frac{b}{c^{2}+1}\ge b-\frac{1}{2}bc[/tex]

[tex]\frac{c}{a^{2}+1}\ge c-\frac{1}{2}ac[/tex]

So
[tex]\frac{a}{b^{2}+1}+\frac{b}{c^{2}+1}+\frac{c}{a^{2}+1}[/tex]

[tex]\ge a-\frac{1}{2}ab+b-\frac{1}{2}bc+c-\frac{1}{2}ac[/tex]

[tex]=a+b+c-\frac{1}{2}(ab+bc+ac)[/tex]

[tex]\ge a+b+c-\frac{1}{2}\cdot\frac{1}{3}(a+b+c)^{2}[/tex] (chebyshev's inequality)

[tex]\ge 3-\frac{1}{2}\cdot\frac{1}{3}(3)^{2}[/tex]

[tex]\ge \frac{3}{2}[/tex]

In the same way, we can prove that the inequalities with five or more variables still hold.

[016hnoor]

Solve the inequality:

[tex](x^2+2x+2)(\sqrt{1-x^2}+2^x)\le \frac{1}{2}[/tex]


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NOOR