Prove Inequality: (ab+cd)^2 ≤ (a^2+c^2)(b^2+d^2)

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The discussion focuses on proving the inequality (ab + cd)² ≤ (a² + c²)(b² + d²). Participants break down the left-hand side (LHS) to ab² + 2abcd + cd² and the right-hand side (RHS) to ab² + (ad)² + (bc)² + cd². The key challenge is to demonstrate that 2abcd ≤ (ad)² + (bc)². Suggestions include rearranging the inequality to form a perfect square, specifically (ad - bc)² ≥ 0, which is always true. The conversation also references the Cauchy–Schwarz inequality for further understanding.
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Homework Statement



Prove that \left(ab+cd\right)^{2} \leq \left(a^{2}+c^{2}\right)\left(b^{2}+d^{2}\right)


Homework Equations



None

The Attempt at a Solution



I've broken the LHS down to the following:

\left(ab\right)^{2}+2abcd+\left(cd\right)^{2}

The RHS:

\left(ab\right)^{2} + \left(ad\right)^{2} + \left(bc\right)^{2} + \left(cd\right)^{2}

So, ultimately... it works out that I need to show 2abcd \leq \left(ad\right)^{2} + \left(bc\right)^{2}

This is where I'm getting stuck... Any suggestions...
 
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rbzima said:

Homework Statement



Prove that \left(ab+cd\right)^{2} \leq \left(a^{2}+c^{2}\right)\left(b^{2}+d^{2}\right)

Homework Equations



None

The Attempt at a Solution



I've broken the LHS down to the following:

\left(ab\right)^{2}+2abcd+\left(cd\right)^{2}

The RHS:

\left(ab\right)^{2} + \left(ad\right)^{2} + \left(bc\right)^{2} + \left(cd\right)^{2}

So, ultimately... it works out that I need to show 2abcd \leq \left(ad\right)^{2} + \left(bc\right)^{2}

This is where I'm getting stuck... Any suggestions...

Yup, so far so good, now subtract 2abcd from both sides, and you'll get:

(ad) ^ 2 - 2(ad)(bc) + (bc) ^ 2 \geq 0

Does the LHS of this inequality remind you of something?
 
VietDao29 said:
Yup, so far so good, now subtract 2abcd from both sides, and you'll get:

(ad) ^ 2 - 2(ad)(bc) + (bc) ^ 2 \geq 0

Does the LHS of this inequality remind you of something?


Wow, long night...
http://scienceblogs.com/insolence/facepalm.jpg​
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Just use (A-B)2=A2-2AB+B2.

Regards.
 
rbzima said:

Homework Statement



Prove that \left(ab+cd\right)^{2} \leq \left(a^{2}+c^{2}\right)\left(b^{2}+d^{2}\right)


Homework Equations



None

The Attempt at a Solution



I've broken the LHS down to the following:

\left(ab\right)^{2}+2abcd+\left(cd\right)^{2}

The RHS:

\left(ab\right)^{2} + \left(ad\right)^{2} + \left(bc\right)^{2} + \left(cd\right)^{2}

So, ultimately... it works out that I need to show 2abcd \leq \left(ad\right)^{2} + \left(bc\right)^{2}

This is where I'm getting stuck... Any suggestions...


Refer to Cauchy–Schwarz inequality for more information (=
 
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Since ##px^9+q## is the factor, then ##x^9=\frac{-q}{p}## will be one of the roots. Let ##f(x)=27x^{18}+bx^9+70##, then: $$27\left(\frac{-q}{p}\right)^2+b\left(\frac{-q}{p}\right)+70=0$$ $$b=27 \frac{q}{p}+70 \frac{p}{q}$$ $$b=\frac{27q^2+70p^2}{pq}$$ From this expression, it looks like there is no greatest value of ##b## because increasing the value of ##p## and ##q## will also increase the value of ##b##. How to find the greatest value of ##b##? Thanks
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