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# Inequality

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This topic contains 2 replies, has 3 voices, and was last updated by  Sonnhard Graubner 8 months, 4 weeks ago.

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• #24712

swastik pramanik
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$$a, b, c>0$$ and $$a+b+c=1$$. Prove that $$\frac{a^2}{b}+\frac{b^2}{c}+\frac{c^2}{a}\geq 3(a^2+b^2+c^2)$$

#24736

Shahbaz Khan
Participant

Use Cauchy-Schwarz on a/b b/c c/a and a,b,c. Should follow easily from this.

#25147

Sonnhard Graubner
Participant

Multiplying by $$abc>0$$ we have to prove that

$$a^3c+ab^3+bc^3\geq 3abc(a^2+b^2+c^2)$$. Since we have

$$a+b+c=1$$ we get

$$(a^3c+ab^3+bc^3)(a+b+c)\geq 3abc(a^2+b^2+c^2)$$

Using the substitution $$b=a+u,c=a+u+v$$ we get

$$4\,{a}^{3}{u}^{2}+4\,{a}^{3}uv+4\,{a}^{3}{v}^{2}+9\,{a}^{2}{u}^{3}+18 \,{a}^{2}{u}^{2}v+15\,{a}^{2}u{v}^{2}+3\,{a}^{2}{v}^{3}+7\,a{u}^{4}+20 \,a{u}^{3}v+21\,a{u}^{2}{v}^{2}+8\,au{v}^{3}+a{v}^{4}+2\,{u}^{5}+7\,{u }^{4}v+9\,{u}^{3}{v}^{2}+5\,{u}^{2}{v}^{3}+u{v}^{4}\geq 0$$ which is true.

Sonnhard.

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