2 renevant inequalities ?

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2 renevant inequalities ?

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#1 h Apr 1, 2025, 11:28 AM

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im confused about 2 inequalities below
1/ Let $a,b,c>0$ . Prove that: $\sum_{cyc} \frac{1+a^2}{1+ab} \geq 3$ 2/ Let $a,b,c>0$ . Prove that: $\sum_{cyc} \frac{1+a^4}{1+ab^3} \geq 3$

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maths_enthusiast_0001

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#2 h Apr 1, 2025, 11:59 AM

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Claim: $\color{blue}\sum_{cyc} \frac{1+a^2}{1+ab} \geq 3$
Proof: We have $a,b,c > 0$ . We wish to prove that,
$\sum_{cyc} \frac{1+a^2}{1+ab} \geq 3$ $\iff \sum_{cyc} (1+a^2)(1+bc)(1+ca) \geq 3(1+ab)(1+bc)(1+ca)$ $\iff \boxed{(a^{3}bc^{2}+a^{2}b^{3}c+ab^{2}c^{3})+(a^{3}c+b^{3}a+c^{3}b)+(a^2+b^2+c^2) \geq (3a^{2}b^{2}c^{2})+abc(a+b+c)+(ab+bc+ca)}$ Now, note that by AM-GM we have,
$(a^{3}bc^{2}+a^{2}b^{3}c+ab^{2}c^{3}) \geq 3(a^{6}b^{6}c^{6})^{1/2}=(3a^2b^2c^2)$ $\implies \boxed{(a^{3}bc^{2}+a^{2}b^{3}c+ab^{2}c^{3}) \geq (3a^2b^2c^2)} \cdots (1)$ Also the well known inequality:
$\boxed{(a^2+b^2+c^2) \geq (ab+bc+ca)} \cdots (2)$ Now by AM-GM we have,
$(a^3c+ab^2c) \geq 2a^2bc$ $(b^3a+abc^2) \geq 2ab^2c$ $(c^3b+a^2bc) \geq 2abc^2$ Adding the above three inequalities we get, $\boxed{(a^3c+b^3a+c^3b) \geq abc(a+b+c)} \cdots (3)$ Adding $(1),(2),(3)$ we get the desired claim. $\blacksquare$

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#3 h Apr 1, 2025, 12:51 PM

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giangtruong13 wrote:

im confused about 2 inequalities below
1/ Let $a,b,c>0$ . Prove that: $\sum_{cyc} \frac{1+a^2}{1+ab} \geq 3$

Let $n\geq 2$ and $x_{1},x_{2},\ldots,x_{n}$ be positive real numbers. Prove that
$\frac{1+x_{1}^2}{1+x_{1}x_{2}}+\frac{1+x_{2}^2}{1+x_{2}x_{3}}+\cdots+\frac{1+x_{n}^2}{1+x_{n}x_{1}}\geq n.$ (All Russian Olympiad 2018)

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#4 h Apr 1, 2025, 2:58 PM

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What about the second one, does anyone have a solution?

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no_room_for_error

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#5 h Apr 1, 2025, 3:02 PM

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giangtruong13 wrote:

What about the second one, does anyone have a solution?

Just AM-GM and then Holder

$(1+a^4)(1+b^4)^3\geq (1+ab^3)^4$

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imnotgoodatmathsorry

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#6 h Apr 8, 2025, 4:16 PM

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giangtruong13 wrote:

im confused about 2 inequalities below
1/ Let $a,b,c>0$ . Prove that: $\sum_{cyc} \frac{1+a^2}{1+ab} \geq 3$ 2/ Let $a,b,c>0$ . Prove that: $\sum_{cyc} \frac{1+a^4}{1+ab^3} \geq 3$

1/
By $AM-GM$ we have: $LHS \ge 3 \sqrt[3]{\frac{(1+a^2)(1+b^2)(1+c^2)}{(1+ab)(1+bc)(1+ca)}}$
So we need to prove: $(1+a^2)(1+b^2)(1+c^2) \ge (1+ab)(1+bc)(1+ca)$ , which is trival because by Cauchy-Schwartz, we have:
$(1+a^2)(1+b^2) \ge (1+ab)^2$ and so on

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#7 h Apr 8, 2025, 4:20 PM • 4 Y

Y by clarkculus, Mogmog8, anduran, pinkdino8074

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