we have to prove that 1/a + 1/b + 1/c >= 3( a+b+c ) multiply both side by ( a+b+c ). Implies, ( a+b+c )( 1/a + 1/b + 1/c ) >= 3*( a+b+c)^2 =3*(a^2+b^2+c^2)+6*(ab+bc+ac) Since ab+bc+ac =1 and a^2 +b^2 +c^2 >= ab+bc+ca We get, ( a+b+c)(1/a+1/b+1/c)>=9. by the converse of of A.M>= H.M. inequality, we proved that.

debasish123 wrote: we have to prove that 1/a + 1/b + 1/c >= 3( a+b+c ) multiply both side by ( a+b+c ). Implies, ( a+b+c )( 1/a + 1/b + 1/c ) >= 3*( a+b+c)^2 =3*(a^2+b^2+c^2)+6*(ab+bc+ac) Since ab+bc+ac =1 and a^2 +b^2 +c^2 >= ab+bc+ca We get, ( a+b+c)(1/a+1/b+1/c)>=9. by the converse of of A.M>= H.M. inequality, we proved that. It's not true

$$LHS=\sum\frac{bc}{a}+2a+2b+2c\geq 3a+3b+3c$$

gobathegreat wrote: Let $a$, $b$ and $c$ be positive real numbers such that $ab+bc+ca=1$. Prove the inequality: $$\frac{1}{a}+\frac{1}{b}+\frac{1}{c} \geq 3(a+b+c)$$ Only use the fact : $$(ab+bc+ca)^2 \ge 3abc(a+b+c)$$$$\Longrightarrow \frac{1}{abc}\ge 3(a+b+c)$$$\blacksquare$

Let $ a $, $ b $ and $ c $ be positive real numbers such that $a^2+b^2+c^2=1$. Prove that$$\frac{1}{a}+\frac{1}{b}+\frac{1}{c} -3(a+b+c)\geq 0$$Let $ a $, $b$ and $c$ be positive real numbers such that $ab+bc+ca=1$. Prove that$$\frac{1}{a^2}+\frac{1}{b^2}+\frac{1}{c^2} -8(a^2+b^2+c^2)\geq 1$$$$\frac{1}{a^3}+\frac{1}{b^3}+\frac{1}{c^3} -15(a^3+b^3+c^3)\geq 4\sqrt 3$$$$\frac{1}{a^4}+\frac{1}{b^4}+\frac{1}{c^4} -26(a^2+b^2+c^2)\geq1 $$Let $ a , b , c>0 $ and $ab+bc+ca+2abc=1$. Prove that $$\frac{1}{a}+\frac{1}{b}+\frac{1}{c} -2(a+b+c)\geq 3$$$$\frac{1}{a^2}+\frac{1}{b^2}+\frac{1}{c^2} -8(a^2+b^2+c^2)\geq 6$$$$\frac{1}{a^3}+\frac{1}{b^3}+\frac{1}{c^3} -16(a^3+b^3+c^3)\geq 18 $$$$\frac{1}{a^4}+\frac{1}{b^4}+\frac{1}{c^4} -32(a^4+b^4+c^4)\geq42 $$

sqing wrote: Let $ a $, $b$ and $c$ be positive real numbers such that $ab+bc+ca=1$. Prove that$$\frac{1}{a}+\frac{1}{b}+\frac{1}{c} -3(a+b+c)\geq 0$$$$\frac{1}{a^4}+\frac{1}{b^4}+\frac{1}{c^4} -26(a^2+b^2+c^2)\geq1 $$

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