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## Main Question or Discussion Point

the problem statement is:

if a,b,c are real numbers such that [tex]\frac{1}{a+1} + \frac{1}{b+1} + \frac{1}{c+1} = 2[/tex]

we have to prove that:

[tex]\frac{1}{4a+1} + \frac{1}{4b+1} + \frac{1}{4c+1} \geq 1[/tex]

thanks in advance.

if a,b,c are real numbers such that [tex]\frac{1}{a+1} + \frac{1}{b+1} + \frac{1}{c+1} = 2[/tex]

we have to prove that:

[tex]\frac{1}{4a+1} + \frac{1}{4b+1} + \frac{1}{4c+1} \geq 1[/tex]

thanks in advance.