7
$\begingroup$

Let $a,b,c,d\in\mathbb R_+$ such that $a+b+c+d\leqslant1$. Prove that$$ \sqrt[4]{\smash[b]{(1-a^4)(1-b^4)(1-c^4)(1-d^4)}}\geqslant255·abcd. $$

My observations:

I can see that all of $a,b,c,d$ are positive fractions which makes all the bracketed factors in LHS positive. Now if we raise both sides to the power $4$, and replace the $4$th powers of $a,b,c,d$ with $A,B,C,D$, our inequality gets reduced to $$(1-A)(1-B)(1-C)(1-D)\ge 255^4 \cdot ABCD$$

Now applying AM $\ge$ GM we get two results,

$$A+B+C+D\ge 4\cdot\sqrt[4]{ABCD}$$

and

$$4-(A+B+C+D)\ge 4\cdot\sqrt[4]{\smash[b]{(1-A)(1-B)(1-C)(1-D)}}$$

Can I somehow use both these results to prove the inequality? Please help

$\endgroup$
8
$\begingroup$

By AM-GM we obtain: $$\sqrt[4]{\prod\limits_{cyc}(1-a^4)}\geq\sqrt[4]{\prod\limits_{cyc}((a+b+c+d)^4-a^4)}=$$ $$=\sqrt[4]{\prod_{cyc}\left((b+c+d)\left((a+b+c+d)^3+(a+b+c+d)^2a+(a+b+c+d)a^2+a^3\right)\right)}\geq$$ $$\geq\sqrt[4]{\prod_{cyc}\left(3\sqrt[3]{bcd}\left(64\sqrt[4]{a^3b^3c^3d^3}+16a\sqrt{abcd}+4\sqrt[4]{abcd}a^2+a^3\right)\right)}\geq$$ $$\geq\sqrt[4]{\prod_{cyc}\left(3\sqrt[3]{bcd}\cdot85\sqrt[85]{(abcd)^{48}\cdot a^{16}\cdot(abcd)^8\cdot(abcd)\cdot a^8\cdot a^3}\right)}=$$ $$=255\sqrt[4]{\prod_{cyc}a^{\frac{84}{85}}b^{\frac{256}{255}}c^{\frac{256}{255}}d^{\frac{256}{255}}}=255abcd.$$

$\endgroup$

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.