# Proof by induction stuck at very end!!

• Nov 24th 2009, 09:20 PM
usagi_killer
Proof by induction stuck at very end!!
• Nov 24th 2009, 11:54 PM
Shanks
It would be better to prove it By using the Second Mathematical Induction(change the inductive hypothesis to $\displaystyle n\le k$)
• Nov 25th 2009, 03:38 AM
dedust
write the inductive hypothesis as
$\displaystyle \left( \sqrt{a_{1}^{2}+b_{1}^{2}}+\sqrt{a_{2}^{2}+b_{2}^{ 2}}+\cdots +\sqrt{a_{k}^{2}+b_{k}^{2}}\right) ^{2}$$\displaystyle \geq \left( \sum\limits_{i=1}^{k}a_{i}^{2}\right) ^{2}+\left( \sum\limits_{i=1}^{k}b_{i}^{2}\right) ^{2} Let\displaystyle RHS denote the right hand side of the above inequality. Notice that \displaystyle \left( \sqrt{a_{1}^{2}+b_{1}^{2}}+\sqrt{a_{2}^{2}+b_{2}^{ 2}}+\cdots +\sqrt{a_{k}^{2}+b_{k}^{2}}+\sqrt{a_{k+1}^{2}+b_{k +1}^{2}}\right) ^{2}$$\displaystyle =\left( \sum\limits_{i=1}^{k}\sqrt{a_{i}^{2}+b_{i}^{2}}\ri ght) ^{2}+2\sqrt{a_{k+1}^{2}+b_{k+1}^{2}}\left( \sum\limits_{i=1}^{k}\sqrt {a_{i}^{2}+b_{i}^{2}}\right) +\left( \sqrt{a_{k+1}^{2}+b_{k+1}^{2}}\right) ^{2}$

$\displaystyle \left( \sum\limits_{i=1}^{k+1}a_{i}\right) ^{2}$$\displaystyle =\left( \sum\limits_{i=1}% ^{k}a_{i}\right) ^{2}+$$\displaystyle 2a_{k+1}\left( \sum\limits_{i=1}^{k}a_{i}\right) +a_{k+1}^{2}$

$\displaystyle \left( \sum\limits_{i=1}^{k+1}b_{i}\right) ^{2}$$\displaystyle =\left( \sum\limits_{i=1}% ^{k}b_{i}\right) ^{2}+$$\displaystyle 2b_{k+1}\left( \sum\limits_{i=1}^{k}b_{i}\right) +b_{k+1}^{2}$

From the inductive hypothesis,

$\displaystyle \left( \sum\limits_{i=1}^{k}\sqrt{a_{i}^{2}+b_{i}^{2}}\ri ght) ^{2}% +2\sqrt{a_{k+1}^{2}+b_{k+1}^{2}}\left( \sum\limits_{i=1}^{k}\sqrt{a_{i}% ^{2}+b_{i}^{2}}\right) +\left( \sqrt{a_{k+1}^{2}+b_{k+1}^{2}}\right) ^{2}$$\displaystyle \geq RHS+2\sqrt{a_{k+1}^{2}+b_{k+1}^{2}}\left( \sum\limits_{i=1}^{k}% \sqrt{a_{i}^{2}+b_{i}^{2}}\right) +\left( \sqrt{a_{k+1}^{2}+b_{k+1}^{2}% }\right) ^{2} and now it only remain to show that \displaystyle 2\sqrt{a_{k+1}^{2}+b_{k+1}^{2}}\left( \sum\limits_{i=1}^{k}\sqrt{a_{i}% ^{2}+b_{i}^{2}}\right) +\left( \sqrt{a_{k+1}^{2}+b_{k+1}^{2}}\right) ^{2}$$\displaystyle \geq2a_{k+1}\left( \sum\limits_{i=1}^{k}a_{i}\right) +a_{k+1}% ^{2}+2b_{k+1}\left( \sum\limits_{i=1}^{k}b_{i}\right) $$\displaystyle +b_{k+1}^{2} Hence, \displaystyle \left( \sqrt{a_{1}^{2}+b_{1}^{2}}+\sqrt{a_{2}^{2}+b_{2}^{ 2}}+\cdots +\sqrt{a_{k+1}^{2}+b_{k+1}^{2}}\right) ^{2}$$\displaystyle \geq \left( \sum\limits_{i=1}^{k+1}a_{i}^{2}\right) ^{2}+\left( \sum\limits_{i=1}^{k+1}b_{i}^{2}\right) ^{2}$