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I am trying to prove the following inequality

For all positive numbers $a$, $b$ and $c$ we have

$$\dfrac{a^3}{b^2-bc+c^2}+\dfrac{b^3}{a^2-ac+c^2}+\dfrac{c^3}{a^2-ab+b^2}\geq a+b+c$$

I can probably solve this by reducing it to Schur's inequality. However, is there any other method?

Lord_Farin
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noot
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4 Answers4

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$\dfrac{a^3}{b^2-bc+c^2}+\dfrac{b^3}{a^2-ac+c^2}+\dfrac{c^3}{a^2-ab+b^2}\geq a+b+c \iff \dfrac{a^3(b+c)}{b^3+c^3}+\dfrac{b^3(a+c)}{a^3+c^3}+\dfrac{c^3(b+a)}{b^3+a^3} \geq a+b+c \iff a^9b+a^9c+b^9a+b^9c+c^9a+c^9b \ge a^7b^3+a^7c^3+b^7a^3+b^7c^3+c^7a^3+c^7b^3$

lemma: $a^n+b^n\ge a^{n-1}b+b^{n-1}a \ge a^{n-2}b^2+b^{n-2}a^2 \ge ...$

$a^n+b^n\ge a^{n-1}b+b^{n-1}a \iff (a-b)(a^{n-1}-b^{n-1})\ge 0 \iff (a-b)^2(a^{n-2}+a^{n-3}b+a^{n-4}b^2+....ab^{n-3}+b^{n-2}) \ge 0$

which is true ,when $a=b$ the "=" holds.

so we have $a^9b+b^9a \ge a^8b^2+a^8b^2 \ge a^7b^3+b^7a^3$

QED.

chenbai
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We need to prove that $$\sum_{cyc}\left(\frac{a^3}{b^2-bc+c^2}-a\right)\geq0$$ or $$\sum_{cyc}\frac{a(a^2+bc-b^2-c^2)}{b^2-bc+c^2}\geq0$$ or $$\sum_{cyc}\frac{a((a-b)(a+2b-c)-(c-a)(a+2c-b))}{b^2-bc+c^2}\geq0$$ or $$\sum_{cyc}(a-b)\left(\frac{a(a+2b-c}{b^2-bc+c^2}-\frac{b(b+2a-c)}{a^2-ac+c^2}\right)\geq0$$ or $$\sum_{cyc}(a-b)^2((a+b)^3-(a^2+3ab+b^2)c+2(a+b)c^2-c^3)(a^2-ab+b^2)\geq0.$$ Now, since by AM-GM $$(a+b)c^2-(a^2+3ab+b^2)c+(a+b)^3\geq2(a+b)^2c-(a^2+3ab+b^2)c>0,$$ it remains to prove that $$\sum_{cyc}(a-b)^2c^2(a+b-c)(a^2-ab+b^2)\geq0.$$ Let $a\geq b\geq c$.

Hence, $$b^2\sum_{cyc}(a-b)^2c^2(a+b-c)(a^2-ab+b^2)\geq$$ $$\geq b^2(a-c)^2b^2(a+c-b)(a^2-ac+c^2)+b^2(b-c)^2a^2(b+c-a)(b^2-bc+c^2)\geq$$ $$\geq a^2(b-c)^2b^2(a+c-b)(a^2-ac+c^2)+b^2(b-c)^2a^2(b+c-a)(b^2-bc+c^2)\geq$$ $$\geq a^2(b-c)^2b^2(a-b)(a^2-ac+c^2)+b^2(b-c)^2a^2(b-a)(b^2-bc+c^2)=$$ $$=(a-b)^2(b-c)^2a^2b^2(a+b-c)\geq0.$$ Done!

Michael Rozenberg
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$$\Longleftrightarrow \dfrac{a^4}{ab^2-abc+ac^2}+\dfrac{b^4}{bc^2-abc+ba^2}+\dfrac{c^4}{ca^2-abc+cb^2}\ge a+b+c$$ by cauchy-Schwarz we have $$\left[ \dfrac{a^4}{ab^2-abc+ac^2}+\dfrac{b^4}{bc^2-abc+ba^2}+\dfrac{c^4}{ca^2-abc+cb^2}\right][bc(b+c)+ac(a+c)+ab(a+b)-3abc]\ge (a^2+b^2+c^2)^2$$ $$\Longleftrightarrow (a^2+b^2+c^2)^2\ge (a+b+c)[(bc(b+c)+ac(a+c)+ab(a+b)-3abc]$$ let $a=\min{\{a,b,c\}}$ $$\Longleftrightarrow a^2(a-b)(a-c)+(b^2+c^2+bc-ab-ac)(b-c)^2\ge 0$$ By Done

math110
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We use the so called Cauchy-reverse technique: \begin{align*} \sum_{\mathrm{cyc}}\frac{a^3}{(b-c)^2+bc}&=\sum_{\mathrm{cyc}}\frac{a^4}{b^2a+c^2a-abc}=a+b+c-\sum_{\mathrm{cyc}}\frac{a^2b^2-a^2bc+a^2c^2-a^4}{b^2a+c^2a-abc}\\ &\ge a+b+c-\sum_{\mathrm{cyc}}\frac{a^2b^2-a^2bc+a^2c^2-a^4}{abc}. \end{align*} It now suffices to show that $\sum_{\mathrm{cyc}} a^2b^2-a^2bc+a^2c^2-a^4\le 0$, or $$a^4+b^4+c^4+abc(a+b+c)\ge 2(a^2b^2+b^2c^2+c^2a^2).$$ However, by Schur's inequality for $r=2$ and AM-GM $$a^4+b^4+c^4+abc(a+b+c)\ge a^3b+a^3c+b^3c+b^3a+c^3a+c^3b\ge 2(a^2b^2+b^2c^2+c^2a^2).$$

Joel Gerlach
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