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if $a,b,c,d$ are positive real numbers,Prove:$$\left(\frac{1}{a}+\frac{1}{b}+\frac{1}{c}+\frac{1}{d}\right)^2\ge\frac{1}{a^2}+\frac{4}{a^2+b^2}+\frac{9}{a^2+b^2+c^2}+\frac{16}{a^2+b^2+c^2+d^2}$$

I was reading the solution of it from book and something was not understandable for me.enter image description hereenter image description here

I have problem in understanding when equality occurs in inequalities that I highlighted them.

for example the first one:$$a^2+c^2 \ge 2 ca$$ $$b^2+c^2\ge 2bc$$ $$a^2+b^2+2c^2 \ge 2(ac+bc)$$ $$4a^2+4b^2+8c^2\ge 8ac+ 8bc$$

but $8a^2+8b^2+8c^2 > 4a^2+4b^2+8c^2$ when $a,b,c$ are positive real numbers.So If i am wrong,where I made mistake?

Also about the third inequality I highlighted,I have problem at proving it.I would appreciate if someone helps me there.

user2838619
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1 Answers1

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For the first:

  • $\displaystyle \frac1{ac+bc}\ge\frac1{a^2+b^2+c^2}$ or $a^2+b^2+c^2\ge ac+bc$

    • By AM_GM: $$\frac{a^2+b^2}2\ge\sqrt{a^2b^2}=ab,\frac{b^2+c^2}2\ge bc,\frac{c^2+a^2}2\ge ca$$
    • Since $ab\ge0$, Adding all: $$a^2+b^2+c^2\ge ab+bc+ca\ge ac+bc$$

For the second:

  • $\displaystyle \frac{4}{b^2+c^2}\ge\frac1{a^2+b^2+c^2}$ or $4a^2+4b^2+4c^2\ge b^2+c^2$ or $4a^2+3b^2+3c^2\ge0$ which is true as $x^2\ge0,\forall x\in\mathbb R$

For the third:

  • $\displaystyle \frac{18}{ab+bd+cd}\ge\frac{16}{a^2+b^2+c^2+d^2}$ or $\displaystyle a^2+b^2+c^2+d^2\ge\frac{8}{9}(ab+bd+cd)$
    • Similiar to first we can prove $a^2+b^2+c^2+d^2\ge ab+bc+cd+da\ge ab+bd+cd\ge \frac{8}{9}(ab+bd+cd)$ since $\frac89<1$
RE60K
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  • thanks for proving the third part.but my Problem is:$4a^2+3b^2+3c^2\ge0$ equality occurs when $a=b=c=0$.but $\frac{1}{a}=\frac{1}{0}$ is undefined.So it should be $4a^2+3b^2+3c^2>0$.there for there is no situation for equality of Left hand side and right hand side. So why $\ge$ have put there? – user2838619 Aug 28 '14 at 08:40
  • try $b=0,c=0,a=1/2$ in $\displaystyle \frac{4}{b^2+c^2}\ge\frac1{a^2+b^2+c^2}$ to yield LHS = RHS – RE60K Aug 28 '14 at 08:52
  • that would result:$\frac{4}{0}\ge\frac{1}{4}$.I think you made a mistake there. – user2838619 Aug 28 '14 at 08:55
  • @user2838619 yes you are correct, anyways if $a>b$ then $a\ge b$ too – RE60K Aug 28 '14 at 08:59