Cool inequality

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Source: Hanoi Specialized School’s Practical Math Entrance Exam (Round 2)

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#1 h Apr 28, 2025, 4:08 PM • 2 Y

Y by Cobedangiu, Exponent11

Let $a,b,c$ be real positive numbers such that: $a^2+b^2+c^2=4abc-1$ . Prove that: $a+b+c \geq \sqrt{abc}+2$

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#2 h Apr 28, 2025, 5:09 PM • 1 Y

Y by Exponent11

Let $abc=t^3$ . By AM-GM, $4t^3-1=a^2+b^2+c^2\ge 3\sqrt[3]{(abc)^2}=3t^2$ , i.e., $(t-1)(4t^2+t+1)\ge 0$ . So, $t\ge 1$ .

Now assume $\sqrt{abc}+2>a+b+c\ge 3t$ . Then $t+2>3t$ , contradicting with $t\ge 1$ .

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#4 h Apr 28, 2025, 5:40 PM

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Please share your ideas with me. You can see in the attachment for the solution. I will recommend you to do inequalities by Sir B.J. Venkatachala

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arqady

#5 h Apr 29, 2025, 7:57 AM

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giangtruong13 wrote:

Let $a,b,c$ be real positive numbers such that: $a^2+b^2+c^2=4abc-1$ . Prove that: $a+b+c \geq \sqrt{abc}+2$

The following inequality a bit of stronger.
Let $a$ , $b$ and $c$ be positive numbers such that $a^2+b^2+c^2=4abc-1$ . Prove that: $a+b+c \geq2\sqrt{abc}+1$

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mudok

#6 h May 4, 2025, 8:57 AM • 1 Y

Y by arqady

arqady wrote:

The following inequality a bit of stronger.
Let $a$ , $b$ and $c$ be positive numbers such that $a^2+b^2+c^2=4abc-1$ . Prove that: $a+b+c \geq2\sqrt{abc}+1$

From the condition $abc\ge 1$

$(a+b+c)^2=4abc-1+2(ab+bc+ca)\ge 4abc-1+2\cdot 3\sqrt[3]{(abc)^2}\ge (2\sqrt{abc}+1)^2$

Done

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