Let $ m,n \in \mathbb{N} $ si $ a,b,c \in \mathbb{R}^*_{+} $ . Prove that:
$$ \displaystyle \frac{a^mb^mc^m(a^n+b^n+c^n)^2}{a^{3m+2n}+b^{3m+2n}+c^{3m+2n}} \le 3 $$
Ineq 2
Re: Ineq 2
What are $\mathbb{N}^*$ and $ \mathbb{NR}^*_{+}$ ?yo79 wrote:Let $ m,n \in \mathbb{N}^* $ si $ a,b,c \in \mathbb{NR}^*_{+} $ . Prove that:
$$ \displaystyle \frac{a^mb^mc^m(a^n+b^n+c^n)^2}{a^{3m+2n}+b^{3m+2n}+c^{3m+2n}} \le 3 $$
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Nur Muhammad Shafiullah | Mahi
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Nur Muhammad Shafiullah | Mahi
Re: Ineq 2
Try using the general mean inequality ($\sqrt[x]{\frac {a^x+b^x+c^x}{3}} \ge \sqrt[y]{\frac {a^y+b^y+c^y}{3}}$ for $x \ge y$) thrice, then multiply them to get the required one.
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Use $L^AT_EX$, It makes our work a lot easier!
Nur Muhammad Shafiullah | Mahi
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Re: Ineq 2
By Chebyshev's inequality,
\[\begin{align*}3(a^{3m}a^{2n}+b^{3m}b^{2n}+c^{3m}c^{2n})&\ge (a^{3m}+b^{3m}+c^{3m})(a^{2n}+b^{2n}+c^{2n})\\
\text{AM/GM}\rightarrow\;&\ge 3a^mb^mc^m(a^na^n+b^nb^n+c^nc^n)\\
\text{Chebyshev}\rightarrow\;&\ge 3a^mb^mc^m\cdot\frac 13(a^n+b^n+c^n)^2.\end{align*}\]
\[\begin{align*}3(a^{3m}a^{2n}+b^{3m}b^{2n}+c^{3m}c^{2n})&\ge (a^{3m}+b^{3m}+c^{3m})(a^{2n}+b^{2n}+c^{2n})\\
\text{AM/GM}\rightarrow\;&\ge 3a^mb^mc^m(a^na^n+b^nb^n+c^nc^n)\\
\text{Chebyshev}\rightarrow\;&\ge 3a^mb^mc^m\cdot\frac 13(a^n+b^n+c^n)^2.\end{align*}\]
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