Show that a/✔️1+a^2+b/✔️1+b^2+c/✔️1+c^2<3✔️3/2 if 1/an1/bc+1/ca=1 through out Jensen 

We are tasked with proving the inequality:

\[
\frac{a}{\sqrt{1 + a^2}} + \frac{b}{\sqrt{1 + b^2}} + \frac{c}{\sqrt{1 + c^2}} < \frac{3\sqrt{3}}{2}
\]

given that:

\[
\frac{1}{a} + \frac{1}{b} + \frac{1}{c} = 1.
\]

### Step 1: Analyze Using Jensen's Inequality

Jensen's inequality applies to convex functions, so the first step is to examine the function \(\frac{x}{\sqrt{1 + x^2}}\).

The second derivative of \( f(x) = \frac{x}{\sqrt{1 + x^2}} \) will help determine whether the function is convex or concave. Let's compute this:

\[
f'(x) = \frac{d}{dx} \left( \frac{x}{\sqrt{1 + x^2}} \right) = \frac{\sqrt{1 + x^2} - \frac{x^2}{\sqrt{1 + x^2}}}{1 + x^2} = \frac{1}{(1 + x^2)^{3/2}}.
\]

Now, the second derivative:

\[
f''(x) = \frac{d}{dx} \left( \frac{1}{(1 + x^2)^{3/2}} \right) = \frac{-3x}{(1 + x^2)^{5/2}}.
\]

Since \( f''(x) < 0 \) for all \( x > 0 \), the function \( f(x) = \frac{x}{\sqrt{1 + x^2}} \) is concave.

### Step 2: Apply Jensen’s Inequality

For concave functions, Jensen’s inequality states that:

\[
f\left( \frac{a + b + c}{3} \right) \geq \frac{f(a) + f(b) + f(c)}{3}.
\]

In our case, the given condition \( \frac{1}{a} + \frac{1}{b} + \frac{1}{c} = 1 \) suggests symmetry among \( a \), \( b \), and \( c \). A natural choice is to set \( a = b = c = \sqrt{3} \), as this satisfies the equation \( \frac{1}{a} + \frac{1}{b} + \frac{1}{c} = 1 \).

### Step 3: Evaluate at \( a = b = c = \sqrt{3} \)

For \( a = b = c = \sqrt{3} \), we calculate:

\[
\frac{\sqrt{3}}{\sqrt{1 + (\sqrt{3})^2}} = \frac{\sqrt{3}}{\sqrt{1 + 3}} = \frac{\sqrt{3}}{2}.
\]

Thus,

\[
\frac{a}{\sqrt{1 + a^2}} + \frac{b}{\sqrt{1 + b^2}} + \frac{c}{\sqrt{1 + c^2}} = 3 \times \frac{\sqrt{3}}{2} = \frac{3\sqrt{3}}{2}.
\]

### Step 4: Conclusion

Since \( f(x) = \frac{x}{\sqrt{1 + x^2}} \) is concave and we have equality for \( a = b = c = \sqrt{3} \), for other values of \( a \), \( b \), and \( c \), we have:

\[
\frac{a}{\sqrt{1 + a^2}} + \frac{b}{\sqrt{1 + b^2}} + \frac{c}{\sqrt{1 + c^2}} < \frac{3\sqrt{3}}{2}.
\]

Thus, we have proved the inequality using Jensen’s inequality.

Would you like further clarification on any part of this explanation?

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**Relative Questions:**
1. What is Jensen’s inequality, and how is it used in inequalities?
2. Why does the concavity of a function matter in Jensen’s inequality?
3. How do you calculate derivatives of functions involving square roots?
4. Can Jensen’s inequality be applied to any concave or convex function?
5. How does symmetry simplify solving inequalities in certain cases?

**Tip:** When solving inequalities with symmetry, try equal values for variables to simplify calculations.

Show that a/✔️1+a^2 + b/✔️1+b^2 + c/✔️1+c^2 < 3✔️3/2 if 1/an + 1/b + 1/c = 1 using Jensen’s Inequality.

Learn how to prove the inequality a/√(1 + a^2) + b/√(1 + b^2) + c/√(1 + c^2) < 3√3/2 using Jensen’s Inequality. This solution covers key mathematical concepts including concavity and inequalities, suitable for advanced high school students.

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