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I was a solving a question on biquadratic equations and almost reached the answer and got a stuck on particular step which had to be solved, I think by AM-GM inequality. The constraints given are as follows:

$2a+b=2$

$0\leqslant a,b\leqslant1$

I had to find to the minimum value of $\,(a^2+b^2)$

I tried to find the value by using AM-GM inequality by splitting $a$ and $b$ in terms of $\frac a2$ or $\frac b2$ etc, but wasn't able to get the desired value of the given function.

I have attached the complete question.

Question

Angelo
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Harshil goyal
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    How did you get 2a+b=2 – Shlok Jain May 26 '23 at 13:24
  • Actually the equation has at least a real solution if and only if $2a+b\geqslant2$. Hence, $$a^2+b^2\geqslant a^2+(2-2a)^2=5a^2-8a+4=\=5\left(a-\frac45\right)^2+\frac45\geqslant\frac45,.$$Also C) is true indeed the area of the graph of $S$ is $\frac14,.$ – Angelo May 26 '23 at 13:33
  • Actually I was just considering the limiting value of the given case. My solution also had the same 2a+b⩾2 – Harshil goyal May 26 '23 at 17:07

3 Answers3

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Since you asked specifically for AM-GM, and even talked about splitting $a$ into $\frac{a}{2}$, you're very close! The answer is Yes it can.

Hint: $ 2a + b = \frac{a}{2} + \frac{a}{2} + \frac{a}{2} + \frac{a}{2} +b $

$$ \frac{2}{5} = \frac{2a+b}{5} = \frac{ \frac{a}{2} + \frac{a}{2} + \frac{a}{2} + \frac{a}{2} +b }{5} \leq \sqrt{ \frac{ \frac{a^2}{4} +\frac{a^2}{4} +\frac{a^2}{4} +\frac{a^2}{4} + b^2 } { 5} } = \sqrt{ \frac{ a^2 + b^2 } { 5}}.$$

Hence, conclude that $a^2 + b^2 \geq \frac{4}{5}$.
Equality occurs when $ \frac{a}{2} = b$ and $2 = 2a+b$.

Calvin Lin
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Another method:

Divide by $x^2$ to get $$x^2+\frac{1}{x^2}+a(x+\frac{1}{x})-b=0$$

Substitute $(x+\frac{1}{x})=t$. So

$f(t)=t^2+at-(b+2)$

it's discriminant is always positive. Now $(x+\frac{1}{x})\geq2$ or $(x+\frac{1}{x})\leq-2$ for all x.

So this equation should have atleast one root greater than 2 or less than -2.

So Either $f(2)<0$ or $f(-2)<0$.

Now f(-2) can never be less than $0$ because a and b are between 0 and 1.

On solving $f(2)>0$ you will get $2a+b\geq2$ (not equal to 2)

So, you have $2a+b\geq2$ and $0<a<1$ and $0<b<1$ if you take a and b on x and y axis respectively. You will get following graph.

graph

Now $a^2+b^2$ represents square of distance of any point in required region from origin. So, it will be minimum when you will draw perpendicular from origin on $2x+y=2$ which is $\frac{2}{\sqrt{5}}$

Square it to get $\frac{4}{5}$ The area of region can also be found

Shlok Jain
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By C-S $$2a+b\leq\sqrt{(4+1)(a^2+b^2)},$$ which gives $$a^2+b^2\geq\frac{4}{5}.$$ The equality occurs for $(a,b)=\left(\frac{4}{5},\frac{2}{5}\right).$

Cauchy-Schwarz(C-S) inequality it's the following.

Let $a_i$ and $b_i$ be real numbers. Prove that: $$(a_1^2+a_2^2+...+a_n)^2(b_1^2+b_2^2+...+b_n)^2\geq(a_1b_1+a_2b_2+...+a_nb_n)^2.$$

In our case $(a_1,a_2)=(a,b)$ and $(b_1,b_2)=(2,1)$.

Michael Rozenberg
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