RMO Practice question 4

Algebra Level 4

( 2 λ − 3 μ ) 2 + ( 2 − 4 λ + 3 μ ) 2 + ( 2 λ − 3 ) 2 + 403 1 2 ≥ η \sqrt{(2\lambda-3\mu)^{2}+ (2-4\lambda+3\mu)^{2}+(2\lambda-3)^{2}+4031^{2}} \geq \eta

Find the maximum integer value of η \eta , such that for all real numbers λ , μ \lambda,\mu , the inequality above is fulfilled.


The answer is 4031.

Shivam Jadhav by Shivam Jadhav , 22, India

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2 solutions

We note that:

( 2 λ − 3 μ ) 2 + ( 2 − 4 λ + 3 μ ) 2 + ( 2 λ − 3 ) 2 + 403 1 2 > 0 + 403 1 2 > 4031 \begin{aligned} \sqrt{(2\lambda-3\mu)^2+(2-4\lambda+3\mu)^2 + (2\lambda-3)^2 +4031^2} & > \sqrt{0 + 4031^2} \\ & > 4031 \end{aligned}

Using Cauchy-Schwartz inequality: [ ( 2 λ − 3 μ ) + ( 2 − 4 λ + 3 μ ) + ( 2 λ − 3 ) ] 2 ≤ 3 [ ( 2 λ − 3 μ ) 2 + ( 2 − 4 λ + 3 μ ) 2 + ( 2 λ − 3 ) 2 ] ⇒ ( − 1 ) 2 ≤ 3 [ ( 2 λ − 3 μ ) 2 + ( 2 − 4 λ + 3 μ ) 2 + ( 2 λ − 3 ) 2 ] ( 2 λ − 3 μ ) 2 + ( 2 − 4 λ + 3 μ ) 2 + ( 2 λ − 3 ) 2 ≥ 1 3 \small [(2\lambda-3\mu)+(2-4\lambda+3\mu) + (2\lambda-3)]^2 \le 3[(2\lambda-3\mu)^2+(2-4\lambda+3\mu)^2 + (2\lambda-3)^2] \\ \Rightarrow (-1)^2 \le 3[(2\lambda-3\mu)^2+(2-4\lambda+3\mu)^2 + (2\lambda-3)^2] \\ (2\lambda-3\mu)^2+(2-4\lambda+3\mu)^2 + (2\lambda-3)^2 \ge \frac{1}{3}

Equality happens when λ = 4 3 \lambda = \frac{4}{3} and μ = 1 \mu = 1 .

Therefore,

( 2 λ − 3 μ ) 2 + ( 2 − 4 λ + 3 μ ) 2 + ( 2 λ − 3 ) 2 + 403 1 2 ≥ 1 3 + 403 1 2 \sqrt{(2\lambda-3\mu)^2+(2-4\lambda+3\mu)^2 + (2\lambda-3)^2 +4031^2} \ge \sqrt{\frac{1}{3} +4031^2}

and the maximum integer value of η \eta is 4031 \boxed{4031} .

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Where did the /4 come from? Shouldn't the answer be 4030?

Calvin Lin Staff - 5 years, 9 months ago

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It is RMS-AM inequality.

Chew-Seong Cheong - 5 years, 9 months ago

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Ah sorry, I didn't notice that you multiplied by 2 later.

However, note that you did not check whether equality can hold. In fact, it cannot, since there is no solution to ( 2 λ − 3 μ ) = ( 2 − 4 λ + 3 μ ) = ( 2 λ − 3 ) = 4031 (2\lambda-3\mu)=(2-4\lambda+3\mu) = (2\lambda-3) =4031 .

In particular, notice that the expression should be at least as large as 403 1 2 = 4031 \sqrt{ 4031^2} = 4031 .

When dealing with inequalities, it is extremely important to verify that the equality case can be achieved. Otherwise, you just have a lower bound, and not the greatest lower bound.

Calvin Lin Staff - 5 years, 9 months ago

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@Calvin Lin – Yes, the minimum value of 403 1 2 \sqrt{4031^2} is obvious.

Chew-Seong Cheong - 5 years, 9 months ago

Sir, is this not correct-

By titu's lemma , we have

( 2 λ − 3 μ ) 2 + ( 2 − 4 λ + 3 μ ) 2 + ( 2 λ − 3 ) 2 + 4031 2 ≥ ( 2 λ − 3 μ + 2 − 4 λ + 3 μ + 2 λ − 3 + 4031 ) 2 4 \large\ { (2\lambda - 3\mu ) }^{ 2 } + { (2 - 4\lambda + 3\mu ) }^{ 2 } + { (2\lambda - 3 })^{ 2 } + { 4031 }^{ 2 } \ge \frac { { (2\lambda - 3\mu + 2 - 4\lambda + 3\mu + 2\lambda - 3{ + 4031) } }^{ 2 } }{ 4 }

and the R.H.S is equivalent to

( 4031 − 1 ) 2 4 \large\ \frac { { ({ 4031 - 1) } }^{ 2 } }{ 4 } .

or equivalent to,

4030 × 4030 4 \large\ \frac { 4030\times 4030 }{ 4 } . . . ( 1 ) ...(1)

Then returning back to the given question we have (from ( 1 ) (1) )

( 2 λ − 3 μ ) 2 + ( 2 − 4 λ + 3 μ ) 2 + ( 2 λ − 3 ) 2 + 4031 2 ≥ ( 2 λ − 3 μ + 2 − 4 λ + 3 μ + 2 λ − 3 + 4031 ) 2 4 \large\ \sqrt { { (2\lambda -3\mu ) }^{ 2 }+{ (2-4\lambda +3\mu ) }^{ 2 }+{ (2\lambda -3 })^{ 2 }+{ 4031 }^{ 2 } } \ge \sqrt { \frac { { (2\lambda -3\mu +2-4\lambda +3\mu +2\lambda -3{ +4031) } }^{ 2 } }{ 4 } }

which is equal to 4030 2 \large\ \frac { 4030 }{ 2 } or 2015 2015 .

This shows that η = 2015 \eta =\boxed{2015} .

Please tell me , where i am wrong.

Priyanshu Mishra - 5 years, 6 months ago

"Mini Sketch of Proof."- Remove the square root in LHS and use partial derivates...

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