Minimum in bi-quadratic!

Algebra Level 5

If a i { a }_{ i } is positive or negative as i i even or odd respectively. If x 4 + a 1 x 3 + a 2 x 2 + a 3 x + 5 = 0 { x }^{ 4 }+{ { a }_{ 1 } }{ x }^{ 3 }+{ a }_{ 2 }{ x }^{ 2 }+{ a }_{ 3 }x+5=0\quad has four Positive roots the them find minimum value of a 1 a 3 { a }_{ 1 }{ a }_{ 3 }

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The answer is 80.

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Deepanshu Gupta by Deepanshu Gupta , 25, India

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2 solutions

Deepanshu Gupta
Oct 6, 2014

Let α , β , γ , δ \alpha ,\beta ,\gamma ,\delta . are 4 positive roots of the given equation.

Now Using Theory of equation:

− a 1 = α + β + γ + δ → ( 1 ) − a 3 = α β γ + β γ δ + γ δ α + δ α β → ( 2 ) { -a }_{ 1 }=\quad \alpha +\beta +\gamma +\delta \quad \quad \quad \quad \quad \quad \quad \quad \quad \quad \quad \quad \quad \quad \quad \quad \rightarrow (1)\\ \\ -{ a }_{ 3 }=\quad \alpha \beta \gamma \quad +\quad \beta \gamma \delta \quad +\quad \gamma \delta \alpha \quad +\quad \delta \alpha \beta \quad \quad \quad \rightarrow (2)\quad .

Also α β γ δ = 5 \alpha \beta \gamma \delta \quad =\quad 5 .

Now using Famous inequality for positive variables

AM ≥ \ge . GM

α + β + γ + δ ≥ 4 α β γ δ 4 α + β + γ + δ ≥ 4 5 4 \alpha +\beta +\gamma +\delta \quad \ge \quad 4\quad \sqrt [ 4 ]{ \alpha \beta \gamma \delta } \\ \alpha +\beta +\gamma +\delta \quad \ge \quad 4\quad \sqrt [ 4 ]{ 5 } \quad .

Similarly

α β γ + β γ δ + γ δ α + δ α β ≥ 4 ( α β γ δ ) 3 4 α β γ + β γ δ + γ δ α + δ α β ≥ 4 ( 5 ) 3 4 \alpha \beta \gamma \quad +\quad \beta \gamma \delta \quad +\quad \gamma \delta \alpha \quad +\quad \delta \alpha \beta \quad \ge \quad 4\quad \sqrt [ 4 ]{ { (\alpha \beta \gamma \delta ) }^{ 3 } } \\ \alpha \beta \gamma \quad +\quad \beta \gamma \delta \quad +\quad \gamma \delta \alpha \quad +\quad \delta \alpha \beta \quad \ge \quad 4\quad \sqrt [ 4 ]{ { (5) }^{ 3 } } .

Now Multiply Two inequalities ( since all terms are Positive)

( α + β + γ + δ ) ( α β γ + β γ δ + γ δ α + δ α β ) ≥ 80 ( − a 1 ) ( − a 3 ) ≥ 80 ( a 1 a 3 ) m i n = 80 (\alpha +\beta +\gamma +\delta )(\alpha \beta \gamma \quad +\quad \beta \gamma \delta \quad +\quad \gamma \delta \alpha \quad +\quad \delta \alpha \beta )\quad \ge \quad 80\\ ({ -a }_{ 1 })(-{ a }_{ 3 })\quad \ge \quad 80\\ { { (a }_{ 1 }{ a }_{ 3 }) }_{ min }\quad =\quad 80 .

Q.E.D

12 Helpful 0 Interesting 0 Brilliant 0 Confused

Sir, actually.. I answered this and got a correct answer of 80 but there is one question in your solution, why multiply two inequalities of same sign, though there is a possible contradiction. I solved this using partial derivatives for f(b, d) = 2bd + (b+d)(sqrt(bd))...

John Ashley Capellan - 6 years, 7 months ago

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No @John Ashley Capellan I already mentioned That all roots are positive So we can multiply inequality and Also I multiply them and said that this a1*a2 is minimum at equality Just only Because While we decide minimum from AM-GM Then at the Minimum

Equality of variable is attained

That means Condition that must be true In inequality - 1 is α = β = γ = δ \alpha =\beta =\gamma =\delta .

and in inequality 2 is

α β γ = β γ δ = α γ δ = α β δ \alpha \beta \gamma =\beta \gamma \delta =\alpha \gamma \delta =\alpha \beta \delta ,

From Both Condition we get Same Unique Value of roots ( For minimum condition )

That is :

α = β = γ = δ = 5 4 \alpha \quad =\quad \beta \quad =\quad \gamma \quad =\quad \delta \quad =\quad \sqrt [ 4 ]{ 5 } \quad .

So in this case we can multiply the inequality.

Hope This May helps You. !!

Deepanshu Gupta - 6 years, 7 months ago

There is mistake in 2nd equation.It should be a3 not a2.

amit kumar - 6 years, 4 months ago

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Thanks ! I fixed That

Deepanshu Gupta - 6 years, 3 months ago

I did same!!

Dev Sharma - 5 years, 6 months ago
Zk Lin
Jan 6, 2016

From Vieta's formula,

− a 1 = α + β + γ + δ -{a}_{1}=\alpha + \beta + \gamma + \delta

− a 3 = α β γ + α β δ + α γ δ + β γ δ -{a}_{3}=\alpha\beta\gamma + \alpha\beta\delta + \alpha\gamma\delta + \beta\gamma\delta

5 = α β γ δ 5= \alpha\beta\gamma\delta

Note that a 1 a 3 {a}_{1}{a}_{3}

= ( α β γ + α β δ + α γ δ + β γ δ ) ( α + β + γ + δ ) =( \alpha\beta\gamma + \alpha\beta\delta + \alpha\gamma\delta + \beta\gamma\delta)(\alpha + \beta + \gamma + \delta)

= ( 5 α + 5 β + 5 γ + 5 δ ) ( α + β + γ + δ ) =(\frac{5}{\alpha}+\frac{5}{\beta}+\frac{5}{\gamma}+\frac{5}{\delta})(\alpha + \beta + \gamma + \delta)

≥ ( 5 + 5 + 5 + 5 ) 2 \geq(\sqrt{5}+\sqrt{5}+\sqrt{5}+\sqrt{5})^{2} by Cauchy-Schwarz inequality

= ( 4 5 ) 2 =(4\sqrt{5})^{2}

= 80 =\boxed{80}

0 Helpful 0 Interesting 0 Brilliant 0 Confused

Moderator note:

What you have demonstrated is a lower bound. You need to explain why this is the greatest lower bound, in order to conclude that we have the minimum value.

For example, it is obvious that a 1 a 3 ≥ 0 a_1 a_ 3 \geq 0 , but 0 is not the answer.

Good point, I always forgot to check equality cases. From the inequality above, suppose that equality holds. Then α 2 = β 2 = γ 2 = δ 2 \alpha^{2}=\beta^{2}=\gamma^{2}=\delta^{2} (conditions for equality cases of Cauchy-Schwarz inequality). Since α , β , γ , δ \alpha,\beta,\gamma,\delta are all positive, we conclude that α 4 = 5 \alpha^{4}=5 or α = β = γ = δ = 5 4 \alpha=\beta=\gamma=\delta=\sqrt[4]{5} , which satisfies the requirement of the problem.

ZK LIn - 5 years, 5 months ago

a 1 ⋅ a 3 > 0 a_1 \cdot a_3 > 0 and not a 1 ⋅ a 3 ≥ 0 a_1 \cdot a_3 ≥ 0

Aditya Sky - 5 years, 2 months ago

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