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Show that for any real number $x$: $$x^2 \sin{x} + x \cos{x} + x^2 + \frac{1}{2} > 0.$$

$\bf{My\; Try::}$ Using $a\sin x+b\cos x\geq -\sqrt{a^2+b^2}$

So $$x^2\sin x+x\cos x\geq -\sqrt{x^4+x^2}=-x\sqrt{1+x^2}$$

and $$4x^4+4x^2+1>4x^4+4x^2\Rightarrow (2x^2+1)^2>4x^2(x^2+1)$$

So $$(2x^2+1)>2x\sqrt{x^2+1}\Rightarrow x^2+\frac{1}{2}>x\sqrt{x^2+1}$$

So $$x^2\sin x+x\cos x+x^2+\frac{1}{2}>-x\sqrt{x^2+1}+x\sqrt{x^2+1}=0$$

So $$x^2\sin x+x\cos x+x^2+\frac{1}{2}\gt 0\;\forall x\; \in \mathbb{R}$$

Is my solution is right, If not Then how can we solve it, Help required, Thanks in Advance

mathlove
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juantheron
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    A general remark about your solution: $\sqrt{x^2} = |x|$ rather than $x$. You claim for example that $-\sqrt{x^4+x^2} = -x\sqrt{1+x^2}$, which is false if $x < 0$. – sTertooy Jul 15 '16 at 13:06

2 Answers2

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Let us check for $$x^2(1+\sin x)+x\cos x-y=0$$

As $x$ is real, the discriminant $$\cos^2x+4(1+\sin x)y\ge0$$

Now $1+\sin x\ge0$

Check what if $1+\sin x=0?$

Else $4y\ge\sin x-1\ge-1-(-1)$

Observe that the equality cannot occur as $1+\sin x\ne0$

lab bhattacharjee
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If we consider $a=(1+\sin x),\, b=\cos(x),\, c=\frac{1}{2}$ we have that $$ b^2-4ac = \cos^2(x)-2-2\sin(x)=-4\sin\left(\frac{\pi}{4}+\frac{x}{2}\right)^4 \leq 0$$ hence $ax^2+bx+c$ is never negative, since $a\geq 0$ and $\Delta=b^2-4ac\leq 0$.

Jack D'Aurizio
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    What about $x^2 \sin{x} + x \cos{x} + x^2 + \frac{1}{2} = 0 $? – Mc Cheng Jul 15 '16 at 13:08
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    @McCheng: that is exactly the question I solved. $a=1+\sin x, ; b=\cos x,; c=\frac{1}{2}$. – Jack D'Aurizio Jul 15 '16 at 13:30
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    In your answer, you have only proved $x^2 \sin{x} + x \cos{x} + x^2 + \frac{1}{2} \ge 0$. You haven't disprove that $x^2 \sin{x} + x \cos{x} + x^2 + \frac{1}{2}$ cannot be equivalence to 0. – Mc Cheng Jul 15 '16 at 13:35
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    What is missing is the case $\Delta = 0$. In that case $x = 2\pi n - \frac{\pi}{2}$ for some $n \in \mathbb{Z}$. But then $\sin(x) = -1$ and $\cos(x) = 0$, so the original equation reduces to $1/2 > 0$ which is clearly true. – sTertooy Jul 15 '16 at 13:47
  • @McCheng: Left to the interested reader, anyway, it just follows from completing the square. – Jack D'Aurizio Jul 15 '16 at 13:47