hajimbrak wrote: Let $a,b$ be positive real numbers.Prove that $(1+a)^{8}+(1+b)^{8}\geq 128ab(a+b)^{2}$. we have : $3a^{4}+b^{4}\geq 4a^{3}b\: \: (1)$ and $3b^{4}+a^{4}\geq 4b^{3}a\: \: (2)$ and $a^{4}+b^{4}\geq 2a^{2}b^{2}\: \: (3)$ from $(1)+(2)+(3)$ we have : $2(a^{4}+b^{4})\geq ab(a+b)^{2}$ and we know $(1+a)^{8}+(1+b)^{8}\geq 256(a^{4}+b^{4})$

It's easy problem. We have by am-gm $ (1+a)^8 \ge 256a^4 $ so we have $ LHS \ge 256(a^4+b^4) $, hence we need to prove that $ 2(a^4+b^4) \ge ab(a+b)^2 $, it's true by chebichev's inequality $ 2(a^4+b^4) \ge (a+b)(a^3+b^3) \ge ab(a+b)(a+b)=ab(a+b)^2 $.

Sardor wrote: hence we need to prove that $ 2(a^4+b^4) \ge ab(a+b)^2 $ Alternatively we can rewrite it as $(a - b)^2 (2 a^2 + 3 a b + 2 b^2) \geq 0$, which is obvious.

$(1+a)^{8}+(1+b)^{8}\geq(2\sqrt{a})^{8}+(2\sqrt{b})^{8}=256(a^4+b^4)$ $\geq 128(a^2+b^2)^2\geq 128ab(a+b)^{2}.$ or $(1+a)^{8}+(1+b)^{8}-128ab(a+b)^{2}\geq(2\sqrt{a})^{8}+(2\sqrt{b})^{8}-128ab(a+b)^{2}$ $=128\left((a^{2}-b^{2})^2+(a-b)^2(a^{2}+ab+b^{2})\right)\geq 0.$

hajimbrak wrote: Let $a,b$ be positive real numbers.Prove that $(1+a)^{8}+(1+b)^{8}\geq 128ab(a+b)^{2}$. There was a stronger version in Kömal's 2014 April issue ($a,b\ge 0$): \[(1+a)^4(1+b)^4\ge 64ab(a+b)^2.\] And here's a generalization ($a,b\ge 0$, $x,y$ real): \[\left(\frac{1+a}2\right)^{2x(x+y)}\left(\frac{1+b}2\right)^{2y(x+y)}\ge a^{x^2}b^{y^2}\left(\frac{a+b}2\right)^{2xy}.\]

randomusername wrote: hajimbrak wrote: Let $a,b$ be positive real numbers.Prove that $(1+a)^{8}+(1+b)^{8}\geq 128ab(a+b)^{2}$. There was a stronger version in Kömal's 2014 April issue ($a,b\ge 0$): \[(1+a)^4(1+b)^4\ge 64ab(a+b)^2.\] And here's a generalization ($a,b\ge 0$, $x,y$ real): \[\left(\frac{1+a}2\right)^{2x(x+y)}\left(\frac{1+b}2\right)^{2y(x+y)}\ge a^{x^2}b^{y^2}\left(\frac{a+b}2\right)^{2xy}.\] Very interesting. Let $a,b$ be positive real numbers.Prove that\[(1+a)^{8}+(1+b)^{8}\geq 2(1+a)^4(1+b)^4\geq128ab(a+b)^{2}.\]

randomusername wrote: There was a stronger version in Kömal's 2014 April issue ($a,b\ge 0$): \[(1+a)^4(1+b)^4\ge 64ab(a+b)^2\] Let $a+b=2u$ and $ab=v^2$, where $v>0$. Hence, $(1+a)^4(1+b)^4\ge 64ab(a+b)^2\Leftrightarrow(1+2u+v^2)^4\geq256u^2v^2\Leftrightarrow$ $\Leftrightarrow1+2u+v^2\geq4\sqrt{uv}\Leftrightarrow2\left(\sqrt{u}-\sqrt{v}\right)^2+(v-1)^2\geq0$, which is obvious.

$(1+a)^8 + (1+b)^8 \geq 2(1+a)^4(1+b)^4 =2((1+ab) + (a+b))^4 \geq 2\left(2\sqrt{(1+ab)(a+b)}\right)^4 = 32(1+ab)^2(a+b)^2 \geq 128ab(a+b)^2$

$a,b$ are positive reals , so let $a=x^2$ and $b=y^2$. $(1+a)^2(1+b)^2=[(1 + x^2)(1+ y^2)]^2=[( x^2 + y^2) + (1 + x^2 y^2)]^2=( x^2 + y^2)^2 + (1+x^2 y^2)^2 + 2(x^2 + y^2)(1 + x^2 y^2) \geq ( x^2 + y^2)^2 + (1+x^2 y^2)^2 + 8xy( x^2 + y^2) \geq 8xy( x^2 + y^2) = 8\sqrt{ab} (a +b) $. So , after squaring , \[(1+a)^4(1+b)^4\ge 64ab(a+b)^2.\]Thus , $(1+a)^{8}+(1+b)^{8}\geq 2(1+a)^4(1+b)^4\ge 128ab(a+b)^2$. QED.

randomusername wrote: There was a stronger version in Kömal's 2014 April issue ($a,b\ge 0$): \[(1+a)^4(1+b)^4\ge 64ab(a+b)^2.\] Cauchy-Schwarz $$(1+a)(1+b)\ge (\sqrt{a}+\sqrt{b})^2$$AM-GM $$(\sqrt{a}+\sqrt{b})^2=a+b+2\sqrt{ab}\ge 2\sqrt{2(a+b)\sqrt{ab}}$$Hence $$(1+a)^4(1+b)^4\ge(\sqrt{a}+\sqrt{b})^8\ge 64ab(a+b)^2$$

cute and fun ^_^

Here we know that 1+a>= 2√a by am gm then we use this and then titu lemma then the inequality become simple